One mole of PCl5  is heated in a closed vessel at volume V litre.  If x moles of PCl5 dissociates into PCl3 and Cl2 at equilibrium, then the expressions for Kc  and KP respectively are (P is the total pressure at equilibrium).

  1. A

    x2(1+x)V,x2P(1+x)

  2. B

    x2(1-x)V,x2P(1-x2)

  3. C

    x2(1+x2)V,x2P(1+x2)

  4. D

    x2(1-x2)V,x2P(1+x2)

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    Solution:

    PCI5PCI3+CI2

      1                0        0                 initially

    1-xv           xv       xv              at equilibrium, Kc=xv,xv(1-xv)=x2(1-x)v

    PCI5PCI3+CI2

      1               0          0                 initially

    1 – x              x           x                  at equilibrium

    1-x1+xp     x1+xp   x1+xp     Partial pressure    :   total number of moles

    = 1 – x + x + x = 1 + x     Partial pressure = mole fraction x total pressure (p)

    Kp=x1+xpx1+xp(1-x1+x)p=Kp=x1+xpx1+xp(1-x1+x)p

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