At temperature T K, PCl5is 50% dissociated at an equilibrium pressure of 4 atm. At what pressure it would dissociate to the extent of 80% at the same temperature:-


a) 0.05 atm
​b) 0.60 atm
c) 0.75 atm
​d) 2.50 atm
​ Kindly answer sir/mam.

The given decomposition reaction is                            PCl5      PCl3    +    Cl2Initial moles         1                   0                 0 At eqm                   1-x             x               xTotal moles = 1-x + x +x = 1 +x Partial press         1-x1+x P     x1 + xP    x1 + xPKP = pPCl3 × pCl2pPCl5      = x1 + xP x1 + xP1-x1+x P  = x2 P12 - x2      = (0.5)2 × 41 - (0.5)2 =1.333      (since x = 50 % dissociation)For 80 % dissociation KP = x2 P12 - x2 1.33 = (0.8)2 × P1- (0.8)2P = 1.33 × 0.360.64 = 0.75 atmThe pressure required for 80 % dissociation = 0.75 atm
Option c is correct

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