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Highly selective Backlog Qs for JEE MainChemistryChemical Equilibrium

The volume percentage of C l 2 at equilibrium in the dissociation of P C l 5 under a total pressure of 1.5 a t m is ( K p = 0.202 ) ,

Options

  1. A74.5
  2. B36.5
  3. C63.5
  4. D26.6

Correct answer

D. 26.6

Step-by-step solution

P C l 5 ⇌ P C l 3 + C l 2 1 - x x x Total moles = 1 - x + x + x = 1 + x k p = P PCl 3 ⋅ PCl 2 P PCl 5 = x 1 + x P 1 − x 1 + x P x 1 + x P k P = x 2 P 1 - x 2 1 - x 2 = 1 k P = x 2 P x = k p P = 0.202 1.5 = 0.134 = 0.366 Mole ∝ volume at constant P & T Volume percentage = m o l e o f C l 2 t o t a l m o l e s × 100 = 0.366 1.366 × 100 = 26.86 %

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