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.

Answer

`3.56xx10^2`

Explanation

`K_c=([NH_3]^2)/([N_2][H_2]^3)`
`=(1.2xx10^(-2))^2/((1.5xx10^(-2))xx(3.0xx10^(-2))^3)`
`=3.56xx10^2`