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Problems · Problem 6.1

Q.The following concentrations were obtained for the formation of NH 3 from N2 and H 2 at equilibrium at 500K. [N2] = 1.5 × 10⁻²M. [H2] = 3.0 × 10⁻² M and [NH3] = 1.2 × 10⁻²M. Calculate equilibrium constant.

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The equilibrium constant KcK_c for the reaction N2+3H2⇌2NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 is found by plugging the given equilibrium concentrations into the law of mass action. The result is Kc=3.56×102 M−2K_c = 3.56 \times 10^2 \, \text{M}^{-2}.

The key idea here is that the equilibrium constant is a fixed number at a given temperature — it tells you the ratio of products to reactants once the reaction has settled. For the formation of ammonia, the balanced equation is:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

The equilibrium constant expression (in terms of concentration, KcK_c) comes directly from the stoichiometry: products over reactants, each raised to the power of its coefficient.

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Notice that the coefficient of NH3\text{NH}_3 is 2, so its concentration is squared. The coefficient of H2\text{H}_2 is 3, so its concentration is cubed. This is not a guess — it’s the law of mass action, and it’s derived from the fact that at equilibrium, the forward and reverse rates are equal.

Now, let’s work through the numbers.

  1. Write down the given concentrations

    [N2]=1.5×10−2 M[\text{N}_2] = 1.5 \times 10^{-2} \, \text{M}

    [H2]=3.0×10−2 M[\text{H}_2] = 3.0 \times 10^{-2} \, \text{M}

    [NH3]=1.2×10−2 M[\text{NH}_3] = 1.2 \times 10^{-2} \, \text{M}

  2. Plug into the KcK_c expression

Kc=(1.2×10−2)2(1.5×10−2)×(3.0×10−2)3K_c = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2}) \times (3.0 \times 10^{-2})^3}

  1. Simplify the numerator

    (1.2×10−2)2=1.44×10−4(1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4}

  2. Simplify the denominator

    First, (3.0×10−2)3=27×10−6=2.7×10−5(3.0 \times 10^{-2})^3 = 27 \times 10^{-6} = 2.7 \times 10^{-5}

    Then multiply by 1.5×10−21.5 \times 10^{-2}:

    (1.5×10−2)×(2.7×10−5)=4.05×10−7(1.5 \times 10^{-2}) \times (2.7 \times 10^{-5}) = 4.05 \times 10^{-7}

  3. Divide numerator by denominator

Kc=1.44×10−44.05×10−7=1.444.05×103K_c = \frac{1.44 \times 10^{-4}}{4.05 \times 10^{-7}} = \frac{1.44}{4.05} \times 10^{3}

1.444.05=0.3555...\frac{1.44}{4.05} = 0.3555..., so Kc≈3.56×102K_c \approx 3.56 \times 10^2

Watch out

A common mistake is to forget the cube on [H2][\text{H}_2] or to square [NH3][\text{NH}_3] incorrectly. Also, note that the units of KcK_c here are M−2\text{M}^{-2} because the denominator has four concentration factors and the numerator has two — so overall, KcK_c has units of inverse molarity squared. In many problems, you can drop units, but it’s good to be aware.

Tip

If you ever get a very large or very small KcK_c, it tells you the position of equilibrium. Here Kc≈356K_c \approx 356, which is moderately large — meaning at 500 K, the formation of ammonia is favoured, but not overwhelmingly so. This matches industrial conditions: ammonia synthesis runs at high pressure to push the equilibrium further toward product.

✓Final answer

The equilibrium constant KcK_c is 3.55×102\boxed{3.55 \times 10^2} (the textbook's rounding; full precision gives 3.56×1023.56 \times 10^2 — in units, M−2\text{M}^{-2}).

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