For a reaction, A(g)⟶B(g) Given, Kp (equilibrium constant) =100 ‌ Temperature ‌=300K ‌ Pressure ‌=1atm Formula used, ∆G∘=−RT‌ln‌Kp . . . (i) Here, ∆G∘= standard Gibb's free energy R= gas constant =8.31Jmol−1K−1 Put value in Eq (i), we get ∆G∘=−R(300)‌ln‌100 ∆G∘=−R(300)(2)‌ln(10) ∵ln(10)=2.3 ∆G∘=−R(300)(2)(2.3) ∆G∘=−1380R ‌ Hence, ‌∆G∘=−xR x=1380