Chemical Equilibrium Chemistry Example 2

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Example 2

medium
For the reaction N2+3H2⇌2NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, at equilibrium [N2]=0.50 M[\text{N}_2] = 0.50\,\text{M}, [H2]=1.50 M[\text{H}_2] = 1.50\,\text{M}, and [NH3]=0.60 M[\text{NH}_3] = 0.60\,\text{M}. Calculate KeqK_{eq}.

Solution

  1. 1
    Keq=[NH3]2[N2][H2]3K_{eq} = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}.
  2. 2
    Substitute: Keq=(0.60)2(0.50)(1.50)3=0.36(0.50)(3.375)K_{eq} = \frac{(0.60)^2}{(0.50)(1.50)^3} = \frac{0.36}{(0.50)(3.375)}.
  3. 3
    Keq=0.361.6875=0.213K_{eq} = \frac{0.36}{1.6875} = 0.213.

Answer

Keq=0.213K_{eq} = 0.213
The equilibrium constant expression uses products over reactants, each raised to the power of their stoichiometric coefficients. A Keq<1K_{eq} < 1 indicates the equilibrium favors reactants.

About Chemical Equilibrium

A dynamic state in a reversible reaction where the forward and reverse reactions proceed at equal rates, so the macroscopic concentrations of reactants and products.

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