`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((2-x))(N_(2))+underset((6-3x))(3H_(2))hArrunderset((2x))(2NH_(3))`
`50%` Dissociation of `N_(2)` take place so,
At equlibrium `(2xx10)/(100)=1,` value of x=1
`K_(c)=([2]^(2))/([1][3]^(3))=4/27so, K_(c)=4/27`