The following concentration were obtained for the formation of `NH_3` from `N_2 and H_2` at equilibrium for the reaction `N_2(g) +3H_2(g) hArr 2NH_3(g)`
`[N_2]=1.5 xx 10^(-2) M`
`[H_2]=3.0 xx 10^(-2) M`
`[NH_3]=1.2 xx 10^(-2) M `
Calculate equilibrium constant.

Explanation

The equilibrium constant for the reaction.
`N(g) + 3H_(2) hArr 2NH_(2)(g)` can be written as
`K_(c ) = ([NH_(3)(g)]^(2))/([N_(2)(g)][H_(2)(g)]^(3))`
`= ((1.2 xx 10^(-2))^(2))/((1.5 xx 10^(-2))(3.0 xx 10^(-2))^(3))`
`= 0.106 xx 10^(4) = 1.06 xx 10^(3)`