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

Question

\(\mathrm{PCl}_{5}\) is \(50 \%\) dissociated at \(20^{\circ} \mathrm{C}\) and 1 atm pressure. The value of \(K_{p}\) is (a) \(0.444\) (b) \(0.555\) (c) \(0.333\) (d) \(0.666\)

Step-by-Step Solution

Answer

The value of \( K_p \) is 0.333.

Step 1: Understand the Reaction and Initial Conditions

The dissociation of phosphorus pentachloride (PCl₅) at equilibrium is represented by the equation: \( \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g) + \mathrm{Cl}_2(g) \). Initially, we start with 1 mole of \( \mathrm{PCl}_5 \) and no \( \mathrm{PCl}_3 \) or \( \mathrm{Cl}_2 \).

Step 2: Find the Equilibrium Concentrations

Since \( \mathrm{PCl}_5 \) is 50% dissociated, at equilibrium, 0.5 moles of \( \mathrm{PCl}_5 \) remain, and 0.5 moles each of \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) are formed. The total moles at equilibrium are \(0.5 + 0.5 + 0.5 = 1.5\).

Step 3: Calculate Partial Pressures

The partial pressures can be calculated using the formula \( P_i = \left( \frac{\text{moles of } i}{\text{total moles}} \right) \times \text{total pressure} \). Therefore, the partial pressures are: - \( P_{\mathrm{PCl}_5} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \) atm- \( P_{\mathrm{PCl}_3} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \) atm- \( P_{\mathrm{Cl}_2} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \) atm

Step 4: Derive the Expression for \(K_p\)

\( K_p \) is given by the formula: \[ K_p = \frac{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} \]Substitute the partial pressures into the expression: \(K_p = \frac{ \left( \frac{1}{3} \right) \times \left( \frac{1}{3} \right) }{ \frac{1}{3} } = \frac{1}{3} \) atm.

Key Concepts

Understanding Dissociation

Dissociation is a chemical process where a compound breaks down into two or more simpler substances. In the context of chemical equilibrium, it's important to understand how dissociation affects the concentrations of products and reactants. For the reaction of phosphorus pentachloride \( \mathrm{PCl}_5 \) dissociating into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \), at 50% dissociation, half of the original \( \mathrm{PCl}_5 \) molecules remain, and the other half is split into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \).This notion of dissociation is essentially observing a balance at the molecular level. - **Why does it matter**: Understanding the extent of dissociation helps predict the concentrations of substances at equilibrium.- **Implications**: The percentage dissociation affects both the equilibrium concentrations and the calculated equilibrium constant.

Equilibrium Constant (\(K_p\))

The equilibrium constant, known as \(K_p\) when using partial pressures, quantifies the ratio of the concentrations of products over reactants at equilibrium, raised to the power of their coefficients in the balanced chemical equation. In the dissociation of \(\mathrm{PCl}_5\), the formula for \(K_p\) is derived from:\[K_p = \frac{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}}\]The partial pressures are plugged into this equation to calculate \(K_p\).- **Importance**: Knowing \(K_p\) provides insight into the degree of reaction completion. A small \(K_p\) suggests the reaction favors reactants, while a larger \(K_p\) indicates product favorability.- **Practical Use**: This constant helps chemists predict and optimize chemical reactions' conditions for desired product yields.

Partial Pressure

Partial pressure refers to the pressure a gas in a mixture would exert if it occupied the entire volume by itself under the same temperature conditions. This concept is crucial when dealing with gases in equilibrium. For the equilibrium mixture of gases in the dissociation of \(\mathrm{PCl}_5\):- The total pressure is the sum of the partial pressures of all gases.- Each component's partial pressure is determined by its mole fraction multiplied by the total pressure.The calculation of partial pressures follows this formula:\[ P_i = \left( \frac{\text{moles of } i}{\text{total moles}} \right) \times \text{total pressure} \]- **Significance**: Knowing the partial pressures allows for precise calculation of \(K_p\). It also helps in understanding how changes in pressure, volume, or temperature can shift the equilibrium.Understanding these fundamental aspects can make mastering chemical equilibrium a much smoother experience.