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
Truong-Son N.
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?

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