The value of equilibrium constant of the reaction, HI `hArr(1)/(2)H_(2) + (1)/(2)I_(2)` is `8` . The equilibrium constant of the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)` will be

Options

A. `1./16`
B. `1//64`
C. 16
D. `1//8`

Answer

B

Explanation

For `HI(g) hArr (1)/(2)H_(2)(g)+(1)/(2) I_(2)(g) ,K=8`
For `(1)/(2)H_(2)(g)+(1)/(2)I_(2)(g)hArr HI(g), K=(1)/(8)`
For `H_(2)(g)+I_(2)(g) hArr 2HI(g)`
`K=((1)/(8))^(2)=(1)/(64)`