Consider the following equilibrium reaction at T K in a 1L flask.
(I) `A(g)iffB(g)+C(g), K_(C_(1))=3`
(II) `B(g)iff D(g)+C(g), K_(C_(2))=?`
If initially 2 moles of A(g) are taken and allowed attain equilibrium, concentration of C(g) was formed to be 3M. What is the value of `K_(C )` of second reaction ?

Options

A. `(200)/(3)`
B. `(266)/(9)`
C. 12
D. `(3)/(200)`

Answer

C

Explanation

`" "A(g) iff B(g) + C(g), K_(C_(1)) = 3`
At equilibrium mole `" "2-x " (x-y) (x+y)"`
`" "B(g) iff D(g) + C(g), K_(C_(2)) = ?`
At equilibrium mole `" "(x-y) " y (x+y)"`
`K_(C_(1)) = 3 = ((x-y)(x+y))/(2-x)" ".....(1)`
`K_(C_(2)) = (y(x+y))/(x-y) " ".....(2)`
`x + y = 3 " "....(3)`
2 - x = x - y
2 = 2x - y
x + y = 3
`x = 5//3, y = 3 - (5)/(3) = (4)/(3)`
`K_(C_(2)) = ((4//3)(3))/((5//3-4//3)))=12`