The value of K_p for the reaction: N₂(g) + 2H₂(g) ⇌ N₂H₄(g) is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Using the equilibrium partial pressures established in the previous sub-question: P_N₂ = 2.5 P₀, P_H₂ = 2.0 P₀, P_N₂H₄ = 1.5 P₀. The equilibrium constant for the second reaction is K_p2 = P_N₂H₄ / (P_N₂ × P_H₂²). Substituting the values: K_p2 = (1.5 P₀) / [2.5 P₀ × (2.0 P₀)²] = 1.5 P₀ / (2.5 P₀ × 4.0 P₀²) = 1.5 / 10 P₀² = 0.15 / P₀² = 3 / (20 P₀²). Therefore, correct answer is D.