For the hypothetic reaction, the equilibrium constant (K) values are given
`A hArr B,K_(1)=2.0 `
`B hArr C, K_(2)=4.0`
`ChArr D, K_(3) =3.0`
The equilibrium constant for the reaction
`A hArr D` is

Options

A. 48
B. 6
C. 2.7
D. 24

Answer

D

Explanation

`K_(1)=([B])/([A])=2.00,K_(2)=([C])/([B])=4.00`
`K_(3)=([D])/([C])=3.00`
For the reaction `A hArr D`
`K=([D])/([A])=([D])/([C])xx([C])/([B])xx([B])/([A])`
`=3xx4xx2=24`