# What is K_(eq) for the reaction N_2 + 3H_2 rightleftharpoons 2NH_3 if the equilibrium concentrations are [NH_3] = 3 M, [N_2] = 2 M, and [H_2] = 1 M?

Use the definition of ${K}_{e q}$.
${K}_{e q} = \left({\left[\text{products"]^(d,e,f, . . . ))/(["reactants}\right]}^{a , b , c , . . .}\right)$
$= \left({\left[{\text{NH"_3]_(eq)^2)/(["H"_2]_(eq)^3["N}}_{2}\right]}_{e q}\right)$
Just make sure you remember to use stoichiometric coefficients correctly. $3 {H}_{2}$ gives ${\left[{\text{H}}_{2}\right]}_{e q}^{3}$ as an equilibrium concentration.