(c) Using equation, ∆G=∆G∘+RT‌ln‌Q where, ∆G→ Change in free energy ∆G∘→ Change in standard free-energy R→ Gas constant T→ Temperature Q→ Reaction quotient At equilibrium, ∆G=0‌ and ‌Q=K So, 0=∆G∘+RT‌ln‌K ⇒∆G∘=−RT‌ln‌K