Question

the K_c for the following reaction at 225 °C is 170. If the equilibrium mixture contains 0.18 M H₂ and 0.020 M N₂, what is the molar concentration of NH₃?

3H₂(g) + N₂(g) ⇄ 2NH₃(g)

[NH₃] = ________ M

Answer 1

K_c = $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

[NH₃]² = K_c [H₂]³ [N₂]

[NH₃]² = 170×[0.18]³×[0.02]

[NH₃]² = 0.01982888

[NH₃] = $\sqrt{0.01982888}$

[NH₃] = 0.14 M