For the following reaction at 50° C and at 2 atm pressure, 2N₂O₅(g) 2N₂O₄(g)+O₂(g) N₂O₅ is 50 % dissociated. The magnitude of standard free energy change at this temperature is x . x= ______ J mol ⁻¹ [Nearest integer]. Given: R=8.314 J mol ⁻¹ K ⁻¹ , 2=0.30 , 3=0.48 , 10=2.303 , °C+273=K

Step-by-step solution

Let the initial moles of N₂O₅ be 2 . The given reaction is: 2N₂O₅(g) 2N₂O₄(g) + O₂(g) Since N₂O₅ is 50 % dissociated (degree of dissociation = 0.5 ), the moles at equilibrium are: Moles of N₂O₅ = 2(1 - 0.5) = 1 Moles of N₂O₄ = 2(0.5) = 1 Moles of O₂ = 0.5 Total moles at equilibrium = 1 + 1 + 0.5 = 2.5 Given the total pressure P = 2 atm, the partial pressures of the gases are: P_ N₂O₅ = 1 2.5 2 = 0.8 atm P_ N₂O₄ = 1 2.5 2 = 0.8 atm P_ O₂ = 0.5 2.5 2 = 0.4 atm The equilibrium constant K_p is: K_p = (P_ N₂O₄ )^2 P_ O₂