Chemqueries: Relationship between Kp ,Kc, Kx and Kn

Sunday, 4 November 2018

Relationship between Kp ,Kc, Kx and Kn

  • Kp=Equilibrium constant in terms of partial pressure.
  • Kc=Equilibrium constant in terms of concentration.
  • Kx=Equilibrium constant in terms of  mole fraction.
  • Kn=Equilibrium constant in terms of  number of moles.
  •  Kp and Kc are related as  KP = KC (RT)∆n ...........................(eq 1)

        Relationship between KP and Kn

            ...........................(eq 2)                                 
        From ideal gas equation,PV=nRT
                                     =>P=n(RT/V)
        Where, n is the number of moles

        So,  PA =nA(RT/v), PB = nB(RT/V), PC =nC(RT/V) and PD =nD(RT/V)
     
        Replacing equation 2 by the above value we get that,     
                 [nC(RT/V)]c  . [nD(RT/V)]d 
     K=  -------------------------------------     
                 [nA(RT/V)]a  . [nB(RT/V)]b
                       nCc . nDd
    = > Kp =  -------------- .  ( RT/V ) (c + d) – (a + b)
                       nAa. nBb  
    = >  Kp = Kn . (RT/V) ∆n

    = >  Kp = Kn . (PT/nT) ∆n                                   

             ∆n=number of gaseous moles of product – number of gaseous moles of reactant

    Relation between  Kp and Kx

           From  above (eq 2)   
                                                         
          Partial pressure(P) = Mole fraction(x) . Total pressure(PT)
          So,   PA =  xA .  PT
                  PB =  xB .  P
                      PC =  xC .  PT
                      PD =  xD .  PT

        Putting the values in whole equation:
                           ( xC .  P)c . ( xD .  PT )d
                 K= ---------------------------------
                           ( xA .  PT )a . ( xB .  PT )b

                             xCc .  xDd
                       =  --------------   PT(c+d) - (a+b)
                             xAa .  xBb

            = >     KP  = Kx  (PT) ∆n 

Q.Check your understanding.
   
      N2O4(g)=2NO2(g) ,Calculate Kand Kfor the given reaction for which Kp=0.157

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