If the value sof `K_(c)` for the reactions, `A+2BhArrC and ChArr2D ` are 2 and 4, respectively, then what will be the value of `K_(c)` for the reaction, `2D hArr A+2B`?

Explanation

`A+2B hArr C,K_(c)=2 =([C])/([A][B]^(2))` . . (1)
`C hArr 2D,K_(c)=4=([D]^(2))/([C])` . . (2)
Adding equations [1] and [2], `A+2BhArrD` . . (3)
The equilibrium constant for equation [3]=product of the equilibrium constants for reactions [1] and [2]
`=([D]^(2))/([A][B]^(2))=2xx4=8`
The equilibrium constant for the reaction `2D hArr A+2B`, the equilibrium constant=reciprocal of the equilibrium constant for reaction `(3)=(1)/(8)=0.125`.