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))hArr2NH_(3(g))`
`[N_(2)]=1.5xx10^(-2)M,[H_(2)]=3.0xx10^(-2)M,[NH_(3)]=1.2xx10^(-2)M`
Calculate the equilibrium constant.

Explanation

`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=(1.2xx10^(-2))/(1.5xx10^(-2)xx(3xx10^(-2))^(3))`
`K_(c)=355.55`