`2` mol of `N_(2)` is mixed with `6` mol of `H_(2)` in a closed vessel of one litre capacity. If `50% N_(2)` is converted into `NH_(3)` at equilibrium, the value of `K_(c)` for the reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`

Options

A. `4//27`
B. `27//4`
C. `1//27`
D. `24`

Answer

A

Explanation

`underset((a-x))oversetaN_(2)+underset((b-3x))overset3H_(2)harrunderset((2x))overset02NH_(3)`
`50%` Dissociation of `N_(2)` take palce so,
At equilibrium `(2xx50)/(100) = 1` , value of x = 1
`K_(c) = ([2]^(2))/([1][3]^(3)) = (4)/(27)`
So `K_(c) = (4)/(27)`c