In a closed flask at 600 K, one mole of X₂ Y₄ (g) attains equilibrium as given below : X ₂ Y ₄(g) 2 XY ₂(g) At equilibrium, 75 % X₂ Y₄ (g) was dissociated and the total pressure is 1 atm. The magnitude of _r G^ (in kJ mol ⁻¹ ) at this temperature is __________. (Nearest Integer) (Given : R = 8.3 J mol ⁻¹ K ⁻¹ ; 10 = 2.3 , 2 = 0.3 , 3 = 0.48 , 5 = 0.69 , 7 = 0.84 )

Step-by-step solution

The given equilibrium reaction is: X ₂ Y ₄(g) 2 XY ₂(g) Initial moles: 1 Degree of dissociation, = 0.75 Moles of X ₂ Y ₄ at equilibrium = 1 - = 1 - 0.75 = 0.25 Moles of XY ₂ at equilibrium = 2 = 2 0.75 = 1.5 Total moles at equilibrium = 0.25 + 1.5 = 1.75 Given total pressure, P = 1 atm. Partial pressure of X ₂ Y ₄ , P_ X ₂ Y ₄ = 0.25 1.75 1 = 1 7 atm Partial pressure of XY ₂ , P_ XY ₂ = 1.5 1.75 1 = 6 7 atm The equilibrium constant K_p is given by: K_p = (P_ XY ₂ )^2 P_ X ₂ Y ₄ = ( 6 7 )^2 1 7 = 36 7 The standard G