The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the 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)`