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ΔG = ΔG° + RTlnQ. At equilibrium ΔG = 0, Q = Keq. So, ΔG° = -RTlnKeq.

1. ΔG° = -8.314 J mol⁻¹ K⁻¹ × 300 K × ln(2 × 10¹³)
2. ΔG° = -2.30RT log K
3. ΔG° = -RTlnKeq
4. ΔG° = RTlnKeq

Correct answer: ΔG° = -RTlnKeq

Solution

At equilibrium, ΔG = 0 and Q = Keq, so the equation simplifies to ΔG° = -RTlnKeq. This is a standard thermodynamic relationship.

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