In a container of constant volume at a particular temparature `N_(2)` and `H_(2)` are mixed in the molar ratio of `9:13`. The following two equilibria are found o be coexisting in the container
`N_(2)(g)+3H_(2)(g)harr2NH_(3)(g)`
`N_(2)(g)+2H_(2)(g)harr N_(2)H_(4)(g)`
The total equiibrium pressure is found to be `305` atm while partial pressure of `NH_(3)(g)` and `H_(2)(g)` are `0.5` atm and `1` atm respectivly. Calculate of equilibrium constants of the two reactions given above.

Explanation

Let the initial partial pressure of `N_(2)` be `9P` and `13P` respectively

Total pressure `=P_(N_(2))+P_(H_(2))+P_(NH_(3))+P_(N_(2)H_(4))=3.5` atm
`(9P-x-y)+(13P-3x-2y)+2x+y=3.5`atm .....(1)
` P_(NH_(3))=2x=0.5` atm .....(2)
`P_(H_(2))=(13P-3x-2y)= 1` atm ....(3)
from (1) `implies(9P-x-y)+1 ` atm `+0.5+y=3.5`
`implies(9P-x)=2` atm
so `9P=2.25`
`P=0.25` atm
from (3) equation `2y=1.5`
`y=0.75` atm
so `P_(N_(2))=9P-x-y=1.25` atm
`P_(H_(2))=1` atm
`P_(NH_(3))=0.5` atm
`P_(N_(2)(H_(4))=0.75` atm
So, `K+(P_(1))=(P_(NH_(3))^(2))/P_(H_(2)^(3).P_(N_(2)))=(0.5xx0.5)/(1xx1xx1xx1.25)=0.2atm^(-2)`
`K_(P_(2)=P_(N_2H_(4))/P_(N_(2).P_(H_(2))^(2))=(0.75)/(1xx1xx1.25)=0.6 atm^(-2)`