The value of equilibrium constant of the reaction, `H(g) hArr(1)/(2) H_(2)(g)` is `0.8` . The equilibrium constant of the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)` will be

Options

A. `(1)/(64)`
B. 16
C. `(1)/(18)`
D. `(1)/(16)`

Answer

A

Explanation

For HI `hArr(1)/(2)H_(2) + (1)/(2)I_(2)` , `K_(C_(1)) = 8`
Reverse given eqn. and multiply by 2.
`:.` For `H_(2) + I_(2)hArr2HI,K_(C_(2)) = ((1)/(K_(c_(1))))^(2)`
`=((1)/(8))^(2) = (1)/(64)`