`A + B hArrC + D`. If finally the concentrations of A an d B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.

Options

A. `4//9`
B. `0*18`
C. `0*17`
D. `0*11`

Answer

D

Explanation

`{:(,A,+,B,hArr,C,D,),(" Intial",a,,a,,,,),(" At eqm",(a-x),,(a-x),,x,,x):}`
But we are given that `x= 2 (a-x)`
or` 3 x = 2 a or a=3/2x = 1.5 x `
` :. K= (x xx x)/(a-x^(2))= x^(2)/(1.5 x - x)^(2) = x^(2)/(0.25 x^(2))=4`