Dissociation of a gas A ₂ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A ₂( ~g ) 2 ~A ( ~g ) The standard Gibbs energy of formation of the involved substances has been provided below: ( array |c|c| Substance & G _ f ^ / kJ ~mol ⁻¹ ~A ₂ & -100.00 ~A & -50.832 array ) The degree of dissociation of A ₂( ~g ) is given by (x 10⁻² )^ 1 / 2 where x= _ _ _ _

Step-by-step solution

For the dissociation reaction A₂(g) 2A(g) , first calculate the standard Gibbs free energy: G°_ rxn = 2(-50.832) - (-100.00) = -1.664 kJ/mol. Using G° = -RT K_p : -1664 = -8 300 K_p , giving K_p = e^ 0.6933 = 2 . For dissociation with degree of dissociation , starting with 1 mole of A₂ at 1 bar total pressure: At equilibrium, moles are (1- ) for A₂ and 2 for A , with total (1+ ) moles. K_p = 4 ^2 1- ^2 = 2 . Solving: 2(1- ^2) = 4 ^2 gives ^2 = 1/3 or = 0.577 . Since degree of dissociation is (x 10⁻²)^ 1/2 : (x 10⁻²