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?

1 Answer
Apr 6, 2017

Use the definition of K_(eq).

K_(eq) = (["products"]^(d,e,f, . . . ))/(["reactants"]^(a,b,c,. . . ))

= (["NH"_3]_(eq)^2)/(["H"_2]_(eq)^3["N"_2]_(eq))

Just make sure you remember to use stoichiometric coefficients correctly. 3H_2 gives ["H"_2]_(eq)^3 as an equilibrium concentration.

Now, plug in your concentrations and evaluate. Are they or are they not already equilibrium concentrations?