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) = (["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.
Now, plug in your concentrations and evaluate. Are they or are they not already equilibrium concentrations?