Question

2 mol of `N_(2)` is mixed with 6 mol of `H_(2)` in a closed vessel of one litre capacity. If `50%` of `N_(2)` is converted into `NH_(3)` at equlibrium, the value of `K_(c)` for the reaction `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` is
A. `4//27`
B. `27//4`
C. `1//27`
D. `24`

Answer 1

Correct Answer - A
`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`