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

Explanation

The equilibrium constant for the reaction. `N_(2)(g) +3H_(2)(g) hArr NH_(3)(g)` can be writtedn as,
`K_(c)=([NH_(3)(g)]^(2))/([N_(2)(g)][H_(2)(g)]^(3))`
`=((1.2xx10^(-2))^(2))/((1.5xx10^(-2))(3.0xx10^(-2))^(3))`
`=0.0318xx10^(4)=3.18xx10^(2)`