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 N₂ is 0.10 atm and H₂ is 0.15 atm?

Question

Quizplus · Accepted Answer

We can use the equilibrium constant expression to solve for the equilibrium partial pressure of ammonia:

KP = (P_NS/P_NH3)^2(P_HS/P_NH3)^3

We can plug in the given values:

1.5 × 10^-13 = (0.10/x)^2(0.15/x)^3

Solving for x, we get:

x = 4.7 × 10^-4 atm

Therefore, the partial pressure of ammonia at equilibrium is 4.7 × 10^-4 atm, which corresponds to choice 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)