For the reaction N₂O₄(g) 2NO₂(g) , the standard free energy change ( G^ ) at 300 K is -1728 J mol ⁻¹ . If the equilibrium is established at a total pressure of 1.5 atm , the percentage dissociation of N₂O₄ is ________. [Given : R = 8.3 J K ⁻¹ mol ⁻¹ , 2 = 0.694 ]

Step-by-step solution

Given, G^ = -1728 J mol ⁻¹ Using the relation G^ = -RT K_p -1728 = -8.3 300 K_p K_p = 1728 2490 0.694 Since 2 = 0.694 , we get K_p = 2 For the equilibrium N₂O₄(g) 2NO₂(g) : K_p = 4 ^2 P 1- ^2 Substitute K_p = 2 and P = 1.5 atm : 2 = 4 ^2 (1.5) 1- ^2 2(1- ^2) = 6 ^2 2 - 2 ^2 = 6 ^2 8 ^2 = 2 ^2 = 0.25 = 0.5 Percentage dissociation = 0.5 100 = 50 % Answer: 50