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 ) ,

Question

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 ) ,, 74.5, 36.5, 63.5, 26.6

MARKS App · Accepted Answer

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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