Question

For the reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the partial pressures of N₂ and H₂ are 0.80 and 0.40 atmosphere respectively at equilibrium. The total pressure of the system is 2.80 atmosphere. What is K_p for the above reaction?

Answer 1

For the reaction, N₂ (g) + 3H₂ (g) ⇌ 2NH₃

Equilibrium constant = Kp = \(\frac{(P_{NH_3})^2}{(P_{N_2})\times(P_{H_2})^3}\)

Given, \(P_{N_2}\) = 0.80 atm

\(P_{N_2}\) = 0.40 atm

P_total = 2.80 atm

\(P_{N_2}\) + \(P_{H_2}\) + \(P_{NH_3}\) = P_total

0.80 + 0.40 + \(P_{NH_3}\) = 2.80

\(P_{NH_3}\) = 2.80 – 1.20

= 1.60 atm

\(\therefore\) K_p = \(\frac{(1.60)^2}{(0.80)\times(0.40)^3}\)

= 50