Question

Thank you!!! ;0;

Answer 1

Answer:

Kc ≅ 100 (1 sig. fig.)

Explanation:

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

C(i)         3.00 mol/L             0                0

ΔC             -2x                       +x              +3x

C(eq)     0.600 mol/L            x                3x

3.00 - 2x = 0.600  =>  x = 3.00 - 0.600/2 = 1.2 mole/L

At equilibrium Kc = [N₂][H₂]³/[NH₃]² = (x)(3x)³/(3·1.2)² = (1.2)(3.6)³/(0.600)² = 155.52 (calculator)

Rounds to 1 sig. fig. based on given 3.00 moles NH₃(g)' Kc ≅ 100.