Question

A+B➡️C+D If initially of A and B are both are taken in equal amount but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction :-

Answer 1

Answer:

$\huge{\sf{\boxed{\boxed{ANSWER}}}}$

Explanation:

Let the volume of the solution or container be = V.    

Initial number of moles of A or B = N.

    One mole of A and one mole of B give rise to one mole of C and one mole of D.  Let there be X moles of C and X moles of D at the equilibrium. So X moles of A and X moles of B are consumed. Remaining amount of A or B will be N-X moles.

$\red{\boxed{\green{A+B=C+D}}}$

       

A + B = C + D

N-X N-X X Xmoles

       

[C] = [D] = X/V ....     [A] = [B] = (N-X)/V 

Given   [D] = 2 [A]

=> X = 2(N-X)   =>   X = 2N/3

   Equilibrium constant = [ C ] [ D ] / {  [ A ] [ B ]  }

                Kc = X² / (N-X)² = 4