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 $[H_2] = 1 M$?

Md Ariful Islam

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Md Ariful Islam

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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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