For the gaseous reaction X(g) 2 Y(g) , the standard Gibbs free energy change G^ is -5727 J mol ⁻¹ at 300 K . A mixture containing 2 moles of X and 3 moles of an inert gas (He) is allowed to reach equilibrium at a constant total pressure of 15 bar at 300 K . The percentage dissociation of gas X at equilibrium is ______. (Given: R = 8.3 J K ⁻¹ mol ⁻¹ , (10) = 2.3 )

Step-by-step solution

First, calculate the equilibrium constant K_p using the relation: G^ = -RT K_p -5727 = -8.3 300 K_p K_p = 5727 2490 = 2.3 Since (10) = 2.3 , K_p = 10 bar . Let be the degree of dissociation of X . X(g) 2 Y(g) Initial moles: 2 0 Equilibrium moles: 2(1- ) 4 Total moles at equilibrium (including 3 moles of He) = 2(1- ) + 4 + 3 = 5 + 2 Partial pressures at equilibrium: p_X = ( 2-2 5+2 ) 15 p_Y = ( 4 5+2 ) 15 The equilibrium constant K_p is: K_p = (p_Y)^2 p_X = ( 4 5+2 15 )^2 2-2 5+2 15 10 = 16 ^2 225 (5+2 )^2 5+2 (2-2