The equilibrium concentration of `NH_(3) , N_(2)` and `H_(2)` are `1.8 xx 10^(-2) M , 1.2 xx 10^(-2)` M and `3 xx 10^(-2) M` respectively . Calculate the equilibrium constant for the formation of `NH_(3)` and `N_(2)` and `H_(2)` . [Hint : M= mol `lit^(-1)`]

Explanation

Given data : `[NH_(3)] = 1. 8 xx 10^(-2) M [N_(2)] = 1.2 xx 10^(-2) M , [H_(2)] = 3 xx 10^(-2) M , K_(C) = ? N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
`K_(C) = ([NH_(3)]^(2))/([N_(2) ] [H_(2)]^(3)) = (1.8 xx 10^(-2) xx 1.8 xx 10^(-2))/(1.2 xx 10^(-2) xx 3 xx 10^(-2) xx 3 xx 10^(-2) xx 3 xx 10^(-2)) = 1 xx 10^(3) L^(2) mol^(-2)`