SHSLSupremeLeader
contestada

Thank you!!! ;0;

3.00 moles of ammonia are originally put into a 1.00-liter container and allowed to decompose (no other chemicals present). At equilibrium, 0.600 moles of ammonia remained in the reaction vessel. What would be the equilibrium constant (K) for this reaction?

2 NH3(g) N2(g) + 3 H2(g(​

Respuesta :

drpelezo

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.

Otras preguntas