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

We can use the equilibrium constant expression as follows:

Kp = (PNS12 * PHS12^3) / PNH3^2

We are given Kp = 1.5 × 10^-13 bar^2 and PNS12 = 0.10 bar and PHS12 = 0.15 bar.

Substituting the values and solving for PNH3, we get:

1.5 × 10^-13 = (0.10 * 0.15^3) / PNH3^2

PNH3^2 = (0.10 * 0.15^3) / 1.5 × 10^-13

PNH3^2 = 1.125 × 10^4

PNH3 = 4.22 × 10^-7 bar

Therefore, the partial pressure of ammonia at equilibrium is 4.7 × 10^-4 bar, which is closest to option B.

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

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