At 298   K , the equilibrium constant of the process 1 . 5 O 2   ( g ) ⇌ O 3   ( g ) is 3 × 10 - 29 . Standard free energy change (in kJ   mol - 1 ) of the process is approximately R = 8 . 314   J   mol - 1   K - 1 ;   log   3 = 0 . 47

Step-by-step solution

Free energy change of a reaction is calculated by, ∆ G = ∆ G ∘ + RT   lnQ At equilibrium, the free energy change of a reaction is zero ( ∆ G = 0 ) , and the reaction quotient ( Q ) is equal to the equilibrium constant ( K ) . So, the above relation takes the form, ∆ G ∘ = - RT   lnK ∆ G ∘ = - 2 . 303 RT   log 10 K where ∆ G ∘ = standard free energy change Therefore, ∆ G ∘ = - 2 . 303 × 8 . 314 × 298 × log 3 ×