In a mixture of `N_(2)` and `H_(2)` in the ratio `1:3` at `30` atm and `300^(@)C`, the `%` of `NH_(3)` at equilibrium is `17.8`. Calculate `K_(p)` for `N_(2)+3H_(2) hArr 2NH_(3)`.

Explanation

Let, the initial moles of `N_2 and H_2` be 1 and 3 respectively (this assumption is valid as `K_p` will not depend on the exact number of moles of `N_2 and H_2` . One can even start with x and 3x )
`{:("",N_2(g),+,3H_2(g),hArr,2NH_3),("Initially",1,"",3,"",0),("At equilibrium",1-x,"",3-3x,"",2x):}`
Since % by volume of a gas is same as % by mole
`therefore (2x)/(4-2x)=0.178`
`therefore x=(4xx0.178)/((2+2xx0.178))=0.302`
`therefore` Mole fraction of `H_2` at equilibrium
`=(3-3x)/(4-2x)=0.6165`
Mole fraction of `N_2` at equilibrium =1-0.6165-0.178 = 0.2055
`therefore K_p=((x_(NH_2) xxP_(T))^2)/((X_(N_2)xxP_T)(X_(H_2)xxP_T)^3)=((0.178xx30)^2)/((0.2055xx30)(0.6165xx30)^3)`
`K_p=7.31 xx 10^(-4) atm^(-2)`