At `400^(@)` and 10 atm pressure, the reaction, `N_2(g)+3H_2(g)to2NH_3(g)` is in a state of equilibrium. If at equilibrium the amount of `NH_3` is 3.85% by volume, then calculate `K_P` for the reaction.

Explanation

Suppose, at equilibrium the extent ofreaction of `N_2=prop`
`N_2(g)+3H_2(g) at equilibrium: `1-prop" "3(1-prop)" "2prop`
Total moles at equilibrium
`=1-prop+3(1-prop)+2prop=2(2-prop)`
mole fraction of `NH_3=(2prop)/(2(2-prop))=(prop)/(2-prop)`=volume
Fraction of `NH_3=0.0385,prop=0.074`
At equilibrium: `n_(N_2)=1-prop=1-0.074=0.926`
`n_(H_2)=3(1-prop)=3(1-0.074)=2.778`
`n_(NH_3)=2prop=0.148`
At equilibrium
`p_(N_2)=x_(N_2)xxp=((0.926)/(0.926+2.7786+0.148))xx10atm=2.4atm`
similarly, `_(H_2)=7.21atm` and `p_(NH_3)=0.384atm`
`K_p=((p_(NH_3)^2)/(p_(N_2))xxp_(H_2)^3)=((0.384)^2)/(2.4xx(7.21)^3)=1.64xx10^(-4)atm^(-2)`