`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. `9//4`
C. `1//9`
D. 4

Answer

D

Explanation

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