The total pressure of gases at equilibrium in the vessel V₂ is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Total mass = 134 g. Molar masses: NH₃(17), N₂(28), H₂(2), Ne(20). Equimolar mixture (x moles each): x(17+28+2+20) = 134 => 67x = 134 => x = 2 moles. Reaction: 2NH₃ ⇌ N₂ + 3H₂. Let 2y moles of NH₃ dissociate. Eq: NH₃ = 2-2y, N₂ = 2+y, H₂ = 2+3y, Ne = 2. N₂ is 52.24% by mass => Mass of N₂ = 0.5224 × 134 = 70 g. Moles of N₂ = 70 / 28 = 2.5. Thus, 2 + y = 2.5 => y = 0.5. Eq moles: NH₃ = 1, N₂ = 2.5, H₂ = 3.5, Ne = 2. Total moles = 9. P_N₂ = 2.5P/9, P_H₂ = 3.5P/9, P_NH₃ = 1P/9. K_p = 2700 atm² (from prior Q). 2700 = [(2.5P/9) × (3.5P/9)³] / (P/9)² = [2.5 × (3.5)³ / 81] × P². P² = (2700 × 81) / (2.5 × (3.5)³). P = √[ (2700 × 81) / (2.5 × (3.5)³) ]. Therefore, correct answer is C.