Question

A+B=C+D if initially of A and B 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

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.

     A    +     B    <===>    C     +      D
   N-X      N-X                 X             X moles

        [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

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We could do quickly using a shortcut,
       as [C] = [D] = 2 [A] = 2 [B],

           Kc = [C] [C] / {[A] [B] } = 2 * 2 = 4

Answer 2

The answer is 4 and the solution is provided in the attachment.
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