Question

A mixture of nitrogen and hydrogen gases, at a temperature of 500

K, was put into an evacuated

vessel of volume 6.0


dm3. The vessel was then sealed.


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


The mixture was allowed to reach equilibrium. It was found that 7.2


mol of N2 and 12.0


mol of H2 were present in the equilibrium mixture. The value of the equilibrium constant, Kc, for this equilibrium is 6.0 × 10–2 at 500

K.


What is the concentration of ammonia present in the equilibrium mixture at 500


K?

answer is B, I don't know why. Please help

Answer 1

0.7589 mol/L ≅ 0.76 mol/L.

Step-by-step explanation:

• For the equilibrium mixture:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g),

Kc = [NH₃]²/[N₂][H₂]³

• At 500 K, V = 6.0 dm³ = 6.0 L:

Kc = 6.0 x 10⁻²,

[N₂] = (7.2 mol)/(6.0 L) = 1.2 mol/L, [H₂] = (12.0 mol)/(6.0 L) = 2.0 mol/L.

∴ [NH₃]² = (Kc)[N₂][H₂]³ = (6.0 x 10⁻²)(1.2 mol/L)(2.0 mol/L)³ = 0.576.

∴ [NH₃] = √(0.576) = 0.7589 mol/L ≅ 0.76 mol/L.