Derive the relation between `K_(p)` and `K_(c)` for the equilibrium reaction.
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`

Explanation

A. `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`K_(p)=(P_(NH_(3))^(2))/(P_(N_(2))xxP_(H_(2))^(3))`
Partial pressures of `P_(N_(2)), P_(H_(2))` and `P_(NH_(3))` can be written as
`P_(N_(2))=(n_(N_(2))RT)/V=C_(N_(2))RT`
`P_(H_(2))=(n_(H_(2))RT)/V=C_(H_(2))RT`
`P_(NH_(3))=(n_(NH_(3))RT)/V=C_(NH_(3))RT`
Substituting these values in `K_(p)` equation
`K_(p)=((C_(NH_(3)))^(2))/((C_(N_(2))(C_(H_(2)))^(3))xx[RT]^(-2)`
`:.K_(p)=K_(c)[RT]^(-2)[ :' (C_(NH_(3))^(2))/((C_(N_(2)))(C_(H_(2)))^(3))=K_(c)]`