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

reactant will be the constant

Answer 2

Answer:  The equilibrium constant of reaction is 4.

Explanation:

                $A+B\rightleftharpoons C+D$

Initially:   x    x            0  0  

At eq'm  x-y    x-y         y    y

Given : [D]=2[A]

i.e y = 2(x-y)

$\frac{y}{2}=x-y$

Equilibrium constant is the ratio of product of concentration of products to the product of the concentration of reactants each term raised to their stochiometric coefficients.

$K_c=\frac{[y][y]}{[x-y][x-y]}$

$K_c=\frac{[y][y]}{\frac{y}{2}\frac{y}{2}}$

$K_c=4$