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)`.

Answer

A, B, D

Explanation

`{:(,N_(2),+,3H_(2),hArr,2NH_(3)),("moles before",1,,3,,0),("dissociation",,,,,),("moles after",(1-x),,(3-3x),,2x),("dissociation",,,,,):}`
`:.` Total moles at equilibrium `=4-2x`
and moles of `NH_(3)` at equilibrium `=2x`
`:' %` of `NH_(3)` in mixture `=17.8`
`:. (2x)/(4-2x)=17.8/100 " " :. X=0.302`
Now, `K_(P)=((n_(NH_(3)))^(2))/(n_(N_(2))xx(n_(H_(2)))^(3))xx(P/(Sigman))^(Deltan)`
`=(4x^(2))/((1-x)(3-3x)^(3))xx(P/(4-2x))^(-2)=(4x^(2)(4-2x)^(2))/(P^(2)(1-x)(3-3x)^(3))`
Now `P=30` atm and `x=0.302`
`:. K_(P)=7.29xx10^(-4) "atm"^(-2)`