For the reaction,
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
the partial pressure of `N_(2)` and `H_(2)` are `0.80` and `0.40` atmosphere, respectively, at equilibrium. The total pressure of the system is `2.80` atm. What is `K_(p)` for the above reaction?

Answer

`K_(p) =50.0 atm^(-2)`

Explanation

The reaction is
`N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g)`
According to avialable data `pN_(2) = 0.80 atm, pH_(2) =0.40 atm , P_("total") =2.8 atm`
`" Thus , "" "pN_(2) +pH_(2) +pNH_(3) =2.8`
`pNH_(3)=2.8 -(pN_(2) +pN_(2)) =2.8 -(0.80 + 0.40) =1.60 atm`
Applying Law of chemical equilibrium :
`K_(p) =(p^(2)NH_(3))/(pN_(2)xxp^(3)H_(2)) =((1.60 " atm" )^(2))/((0.80" atm")xx(0.4" atm")^(3)) =50.0 " atm"^(-2)`