Consider the equilibrium: N ₂ O ₄( g ) 2 NO ₂( g ) At 300 K and 1 atm total pressure, the vapor density of the equilibrium mixture is 34.5 . The standard free energy change G^ for the reaction at this temperature is ________ J mol ⁻¹ . (Nearest integer) [Given: 2 = 0.693 and R = 8.3 J K ⁻¹ mol ⁻¹ ]

Step-by-step solution

The molar mass of N ₂ O ₄ is 2(14) + 4(16) = 92 g mol ⁻¹ . The vapor density of pure N ₂ O ₄ ( D ) is 92 2 = 46 . The degree of dissociation is related to the vapor density of the mixture ( d ) by: = D - d d(n - 1) Here, 1 mole of N ₂ O ₄ gives 2 moles of NO ₂ , so n = 2 . = 46 - 34.5 34.5(2 - 1) = 11.5 34.5 = 1 3 For the reaction N ₂ O ₄( g ) 2 NO ₂( g ) : Initial moles: 1 0 Equilibrium moles: 1- 2 Total moles at equilibrium = 1 + = 1 + 1 3 = 4 3 Partial pressures at P = 1 atm : P_ N ₂ O ₄ = 1- 1+ P = 2/3 4/3 1 =