In a container of constant volume at a particular temperature `N_2 and H_2` are mixed in the molar ratio of 9:13. The following two equilibria are found to be coexisting in the container
`N_2(g)+3H_2(g)hArr2NH_3(g)`
`N_2(g)+2H_2(g)hArrN_2H_4(g)`
The total equilibrium pressure is found to be 3.5 atm while partial pressure of `NH_3(g) and H_2(g)` are 0.5 atm. and 1 atm respectively.Calculate of equilibrium constants of the two reactions given above.Give your answer as `(Kp_1 + Kp_2)xx10`

Answer

8

Explanation

Let the initial partial pressures of `N_2` and `H_2` be 9P and 13 P respectively.
`{:(N_2(g)+,3H_2(g)hArr,2NH_3(g)),(2P-y-x,13P-3x-2y,2x):}`
`{:(N_2(g)+,2H_2(g)hArr,N_2H_4(g)),(2P-y-x,13P-3x-2y,y):}`
Total pressure=`P_(N_2)+P_(H_2)+P_(NH_3)+P_(N_2H_4)`=3.5 atm
=(9P-x-y)+(13 P-3x-2y)+2x+y=3.5 atm ...(i)
`P_(NH_3)=2 x=0.5` atm ...(2)
`P_(H_2)=(13P-3x-2y)=1` atm
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)`=9P-x-y=1.25 atm
`P_(H_2)`=1atm
`P_(NH_3)`=0.5 atm
`P_(H_2H_4)=`0.75 atm
So,
`K_(P_1)=(P_(NH_3)^2)/(P_(H_2)^3.P_(N_2^2))=(0.5xx0.5)/(1xx1xx1xx1.25)=0.2 atm^(-2)`
`K_(P_2)=(P_(N_2H_2))/(P_(N_2).P_(H_2^2))=(0.75)/(1xx1xx1.25)=0.6 atm^(-2)`
`P_(A_2C_2)`=y-z=1-`1/4=3/4` atm