Suppose that you are running this reaction in a sealed flask. N2 (g) + 3 H2 (g) <--> 2 NH3 (g)

Question

If [N2] = 1.4 M, [H2] = 0.80 M, and [NH3] = 0.35 M, what is Keq under the conditions of this flask?

Answer 1

You can't answer this question unless you know those concentrations prevail at equilibrium. If at equilibrium, then
Keq = (NH3)^2/(N2)(H2)^3
Plug in the numbers and turn the crank.