The decomposition of ammonia is 2NH₃(g) → N₂(g) + 3H₂(g). If K_p is 1.5 × 10³ bar² at 400 °C, what is the partial pressure of ammonia at equilibrium when the partial pressure of N₂(g) is 0. 10 bar and that of H₂(g) is 0.15 bar?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: 
KS1U1B1pS1U1B0 = [NS1U1B12S1U1B0] [HS1U1B12S1U1B0]^3 / [NH3]^2 
At equilibrium, [NS1U1B12S1U1B0] = 0.10 bar and [HS1U1B12S1U1B0] = 0.15 bar. 
Substituting these values and solving for [NH3], we get: 
1.5 × 10S1U1P13S1S1P0 barS1U1P12S1S1P0 = (0.10) (0.15)^3 / [NH3]^2 
[NH3]^2 = (0.10) (0.15)^3 / (1.5 × 10S1U1P13S1S1P0 barS1U1P12S1S1P0) 
[NH3]^2 = 4.6875 × 10S1U1P-2S1S1P0 
[NH3] = 2.165 × 10S1U1P-1S1S1P0 bar (approx.) 
Therefore, the partial pressure of ammonia at equilibrium is 2.2 × 10S1U1P1-S1S1P0S1U1P14S1S1P0 bar (rounded to 2 significant figures).

The Decomposition of Ammonia Is 2NH3(g) → N2(g) + 3H2(g)

The Decomposition of Ammonia Is 2NH₃(g) → N₂(g) + 3H₂(g)