Question

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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(​

Answer 1

Kc ≅ 100 (1 sig. fig.)

Step-by-step 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.