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.
A. `1.08 xx 10^(4)`
B. `3.98 xx 10^(2)`
C. `1.06 xx 10^(3)`
D. `2.93 xx 10^(4)`