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. `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`

Explanation

`{:("Reaction:" ,N_(2(g)) + , 3H_(2(g)) hArr , 2NH_(3(g))),("Concentration at equilibrium:" , 1.5xx10^(-2)M, 3.0xx10^(-2)M, 1.2xx10^(-2)M):}`
The equilibrium constant is `K_c` ,
`K_c=([NH_3]^2)/([N_2][H_2])^3`
`=(1.2xx10^(-2) M)^2/((1.5xx10^(-2)M)(3.0xx10^(-2)M)^3)`
`=(1.2)^2/((1.5)(3.0))^3 xx (10^(-2))^2/((10^(-2))xx(10^(-2))^3)M^((2-3-1))`
`=3.556xx10^(-2)xx10^4`
`=3.556xx10^2 M^(-2)`