Answer to Question #310111 in Chemistry for Lei

Question

Consider the chemical reaction 2NH₃(g) ⇌ N₂(g) + 3H₂(g). The equilibrium is to be established in a 1.0 L container at 1,000 K, where K_c = 4.0 × 10^–2. Initially, 1,220 moles of NH₃(g) are present. Estimate the equilibrium concentration of N₂(g). 

Answer 1

Solution:

Molarity of NH₃ = Moles of NH₃ / Volume of container = (1220 mol) / (1.0 L) = 1220 M

Balanced chemical equation:

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

From this, we can construct an ICE table:

The K_c expression for the reaction is:

From the ICE Table:

[NH₃] = 1220 − 2X ≈ 1220 (since 1220 >> 2X)

[N₂] = X

[H₂] = 3X

K_c = 4.0×10^–2 (according to the task)

Therefore,

(4.0×10^–2) × (1488400) = 27X⁴

59536 = 27X⁴

X⁴ = 2205.04

X = (2205.04)^1/4 = 6.85

Therefore,

[NH₃] = (1220 − 2X) M = (1120 − 2×6.85) M = 1206.3 M

[H₂] = (3X) M = (3×6.85) M = 20.55 M

[N₂] = X = 6.85 M

Answer: The equilibrium concentration of N₂(g) is 6.85 M