Question

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`?

Answer 1

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?