Question

The degree of dissociation of PCl₅(g) obeying the equilibrium PCl₅ rightleftharpoons PCl₃ + Cl₂ is related to the equilibrium pressure P by:

• A. \(\alpha \propto \dfrac{1}{P^4}\)
• B. \(\alpha \propto \dfrac{1}{\sqrt{P}}\)
• C. \(\alpha \propto \dfrac{1}{P^2}\)
• D. α ∝ P

Hint:

For dissociation reactions: Kp ≈ α²P when α ≪ 1.

Answer 1

The Correct Option is B

Step 1: Let initial moles of PCl₅=1, degree of dissociation =α.
Step 2: Equilibrium moles: PCl₅=1-α, PCl₃=α, Cl₂=α
Step 3: Equilibrium constant: Kp=(α²P)/((1-α)P) ≈ α²P
Step 4: α ∝ \frac1√(P)