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Two moles of P C l 5 were heated in a closed vessel of 2 L. At equilibrium 40% of P C l 5 is dissociated into P C l 3 a n d C l 2 . The value of equilibrium constant is

Options

  1. A0.53
  2. B0.267
  3. C2.63
  4. D5.3

Correct answer

B. 0.267

Step-by-step solution

Degree of dissociation=0.4 PCl 5 ⇌ PCl 3 + Cl 2 a ( 1 − x ) a x a x a = 2 , x = 0.4 , V = 2 P C l 5 = 2 ( 1 - 0.4 ) 2 = 0.6 mol/L P C l 3 = 2 × 0.4 2 = 0.4 mol/L C l 2 = 2 × 0.4 2 = 0.4 mol/L ∴ K c = P C l 3 C l 2 P C l 5 = 0.4 × 0.4 0.6 = 0.267

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