Two moles of `NH_(3)` when put into a proviously evacuated vessel (one litre) pertially dissociate into `N_(2)` and `H_(2)`. If at equilibrium one mole of `NH_(3)` is present, the equilibrium constant is

Options

A. `3//4 mol^(2)litre^(-2)`
B. `27//64 mol^(2)litre^(-2)`
C. `27//32 mol^(2)litre^(-2)`
D. `27//1 mol^(2)litre^(-2)`

Answer

D

Explanation

`{:(,2NH_(3),hArr,N_(2),+,3H_(2)),(underset("at.equal")("initial"),2,,0,,0),(,1,,1,,3):}`
`K = ([N_(2)][H_(2)]^(3))/([NH_(3)]^(2)) = (1xx3^(3))/(1) = 27`