Chemqueries: Equilibrium constant in terms of partial pressure.

Friday, 2 November 2018

Equilibrium constant in terms of partial pressure.

 Both KP and Kc  are equilibrium constants,but  Kc  is defined by molar concentration   

but  KP is defined by partial pressure.

Derivation of KP :

  • Consider the following reversible reaction

                 aA(g) + bB(g) ⇔ cC(g) + dD(g) 

  • If equilibrium constant for the reaction  is expressed in terms of the molar concentration ,
            Equilibrium Constant

  • If the equilibrium involves gaseous species, then concentration may be expressed in terms of partial pressure of the reactants and products and it is represented as

                 ......................... (eq1)
                                                                                                                                                           
PA  ,  PB  ,  PC and PD  are  the partial pressure of A,B,C and D respectively.

Relationship between  KP and Kc  :

We know that ,for a ideal gas pressure is directly proportional to the concentration assuming volume and temperature constant.
   So,    PV = nRT
   Divide both sides by V,
            P=(n/V)RT
            P=CRT
  Where,C is concentration
               n is the number of moles
               V is the volume
                R is ideal gas constant,and
                n/V is the molar concentration.
Thus the partial pressures of individual gases A, B, C and D are:
PA = [A]RT; PB = [B]RT; PC = [C]RT; PD = [D]RT
Substituting these value in equilibrium constant expression equation 1, we have



    KP = KC (RT)(c+d)-(a+b)

    KP = KC (RT)∆n   
where ∆n = (c+d) – (a+b)
                =number or gaseous moles of the product-number of gaseous moles of the reactant






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