At 525 K, PCl₅(g) is 80% dissociated at a pressure of 1 atm. Now, sufficient quantity of an inert gas at constant pressure is introduced into the above reaction mixture to produce inert gas partial pr

Question

At 525 K, PCl₅(g) is 80% dissociated at a pressure of 1 atm. Now, sufficient quantity of an inert gas at constant pressure is introduced into the above reaction mixture to produce inert gas partial pressure of 0.9 atm. What is the percentage dissociation of PCl₅(g) when equilibrium is re-established?

Prashant Kotian · Accepted Answer

Correct answer: A. For PCl₅ ⇌ PCl₃ + Cl₂, initial K_p = [0.8² × 1] / [1 - 0.8²] = 0.64 / 0.36 = 16/9. With inert gas, partial pressure of reacting system P' = 1 - 0.9 = 0.1 atm. New eq: [α² × 0.1] / [1 - α²] = 16/9. Solving gives 0.9α² = 16 - 16α² => 16.9α² = 16 => α ≈ 0.973 = 97.3%. Therefore, correct answer is A.

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