A rigid flask of volume 16.4 L contains 3.0 moles of N ₂ O ₄ and an unknown amount of Helium gas at 400 K. The system is allowed to reach equilibrium according to the reaction: N ₂ O ₄( g ) 2 NO ₂( g ) At equilibrium, the total pressure of the gaseous mixture is found to be 14.0 atm. If the equilibrium constant K_p for the reaction is 4.0 atm, the number of moles of Helium gas initially added is: [Given: R = 0.082 L

Step-by-step solution

The pressure exerted per mole of gas in the flask is given by: P n = RT V = 0.082 400 16.4 = 2.0 atm mol ⁻¹ Initial partial pressure of N ₂ O ₄ = 3.0 2.0 = 6.0 atm. Let p be the decrease in pressure of N ₂ O ₄ at equilibrium. N ₂ O ₄( g ) 2 NO ₂( g ) Initial: 6.0 atm 0 Equilibrium: 6.0 - p 2p The equilibrium constant K_p is given by: K_p = (P_ NO ₂ )^2 P_ N ₂ O ₄ = (2p)^2 6.0 - p = 4.0 4p^2 = 24 - 4p p^2 + p - 6 = 0 (p+3)(p-2) = 0 Since p must be positive, p = 2.0 atm. At equilibrium, the partial pressures are: P_