The decomposition of ammonia is 2 NH₃(g)→ N₂(g)+ 3 H₂(g).If K_p is 1.5 × 10³ 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

First, we need to write the expression for the equilibrium constant:
Kp = (P_NS12 * P_H2S3) / P_NH3^2
Since we are given the values for P_NS12 and P_H2S3, we can substitute these into the equation and solve for the partial pressure of ammonia:
1.5 × 10^13 = (0.10 * 0.15^3) / P_NH3^2
P_NH3^2 = (0.10 * 0.15^3) / 1.5 × 10^13
P_NH3 = sqrt((0.10 * 0.15^3) / 1.5 × 10^13)
P_NH3 = 4.7 × 10^-14
Therefore, the partial pressure of ammonia at equilibrium is 4.7 × 10^-14 bar, which corresponds to option B.

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

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