Question

\(PCl_{5}\) is dissociating 50% at 250°C at a total pressure of \( P \) atm. If the equilibrium constant is \( K_p \), then which of the following relation is numerically correct?

• A. \( K_p=3P \)
• B. \( P = 3K_p \)
• C. \( P = \frac{2K_p}{3} \)
• D. \( K_p = \frac{2P}{3} \)

Hint:

For dissociation reactions, the equilibrium constant \( K_p \) can be calculated using the partial pressures of the reactants and products. Ensure that the stoichiometry of the reaction is correctly accounted for in the calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Dissociation of \(\text{PCl}_5\) 
The dissociation reaction of phosphorus pentachloride is: \[ \text{PCl}_5 (g) \leftrightarrow \text{PCl}_3 (g) + \text{Cl}_2 (g). \] At 50% dissociation, half of the \(\text{PCl}_5\) dissociates into \(\text{PCl}_3\) and \(\text{Cl}_2\). 

Step 2: Calculating Partial Pressures
 Let the initial moles of \(\text{PCl}_5\) be 1. At 50% dissociation: \[ \text{Moles of PCl}_5 = 0.5, \quad \text{Moles of PCl}_3 = 0.5, \quad \text{Moles of Cl}_2 = 0.5. \] The total moles at equilibrium are: \[ 0.5 + 0.5 + 0.5 = 1.5. \] The partial pressures are: \[ P_{\text{PCl}_5} = \frac{0.5}{1.5} P = \frac{P}{3}, \quad P_{\text{PCl}_3} = \frac{0.5}{1.5} P = \frac{P}{3}, \quad P_{\text{Cl}_2} = \frac{0.5}{1.5} P = \frac{P}{3}. \] 

Step 3: Calculating the Equilibrium Constant \(K_p\) 
The equilibrium constant \(K_p\) is given by: \[ K_p = \frac{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}}{P_{\text{PCl}_5}}. \] Substituting the partial pressures: \[ K_p = \frac{\left(\frac{P}{3}\right) \cdot \left(\frac{P}{3}\right)}{\left(\frac{P}{3}\right)} \] \[ = \frac{\frac{P^2}{9}}{\frac{P}{3}} = \frac{P}{3}. \] Therefore, the correct relation is: \[ P = 3K_p. \] 

Step 4: Matching with the Options 
The correct relation \( P = 3K_p \) corresponds to option (B).

Final Answer: The numerically correct relation is (B) \( P = 3K_p \).