`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+,BhArr,C+,D),("Initial moles",a,a,0,0),("Equilibrium moles",a-x,a-x,x,x):}`
At equilibrium, the concentration of D is twice that of A. Thus,
`x=2(a-x)`
or `3x=2a`
or `a= (3)/(2)x= 1.5x`
According to the of chemical equilibrium,
`K_(c)=(C_(C).C_(D))/(C_(A).C_(B))`
`= ((x)(x))/((a-x)(a-x))=(x^(2))/((a-x)^(2))`
Substituting the value of a, we get
`K_(c)= (x^(2))/((1.5x-x)^(2))`
`= (x^(2))/([x(1.5-1)]^(2))`
`= (x^(2))/(0.25x^(2))=4`