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//16 "mol"^2 "litre"^(-2)`

Answer

D

Explanation

`2NH_3 hArr N_2 +3H_2`
`{:("Initial concentration",2,0,0),("At equilibrium concentration",1,1//2,3//2):}`
`K_c = ([N_2]\[H_2]^3)/([NH_3]^2)=(1//2xx[3//2]^3)/(1^2)=27/16`
[Given `2 - 2x = 1 , 2x = 1 , x = 1//2` ]