If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`

Options

A. `7`
B. `(1)/(7)`
C. `sqrt((1)/(7))`
D. `49`

Answer

A

Explanation

`H_(2)(g)+l_(2)(g)hArr2HI(g)`
`K_(c)=([HI]^(2))/([H_(2)][l_(2)]) "if" [H_(2)]=[l_(2)]`
`K_(c)=([HI]^(2))/([l_(2)]^(2)) [HI]^(2)=K_(c)xx[l_(2)]`
or `([HI]^(2))/([l_(2)]^(2))=K_(c) "or" ([HI])/([l_(2]])=sqrtK_(c)=sqrt49=7`