Question

The Kc for the following reaction at 225 ∘C is 1.7×102. 3H2(g)+N2(g)⇌2NH3(g) If the equilibrium mixture contains 0.23 M H2 and 0.023 M N2, what is the molar concentration of NH3?

Answer 1

Answer: 0.2197 M

Step-by-step explanation:

3H2(g) + N2(g) = 2NH3(g)

Kc = [NH3]/[H2]³[N2]

[NH3]² = Kc x [H2]³ x [N2]

[NH3] = √Kc x [H2]³ x [N2]

Since Kc = 1.7x10², H2=0.23M, N2=0.023M, NH3=?

[NH3] = √1.72x10² x (0.23)³ x (0.023)

[NH3] = √0.0483

[NH3] = 0.2197M