For the following aqueous reaction at 298 K: N2(g) + 3 H2(g) ⇌ 2 NH3(g), what is the value of the equilibrium constant K if the equilibrium concentrations are [N2] = 0.20 M, [H2] = 0.30 M, and [NH3] = 0.40 M?

• A. K = 44.4
• B. K = 0.15
• C. K = 0.037
• D. K = 2.22

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation: \(\mathrm{N_2(g) + 3\ H_2(g) \rightleftharpoons 2\ NH_3(g)}\).
2. Recall the expression for the equilibrium constant \(K\) in terms of concentrations: \(K = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\).
3. Substitute the given equilibrium concentrations into the expression: \(K = \frac{(0.40)^2}{(0.20)(0.30)^3}\).
4. Calculate the numerator by squaring the concentration of \(\mathrm{NH_3}\): \((0.40)^2\).
5. Calculate the denominator by multiplying the concentration of \(\mathrm{N_2}\) by the cube of the concentration of \(\mathrm{H_2}\): \((0.20) \times (0.30)^3\).