Question Detail
Question
For the reaction A → B the standard free energy change ΔG° = -5.0 kJ/mol at 298 K. What is the equilibrium constant K for the reaction? R = 8.314 J/mol/K
Correct Answer
Option A is the correct answer.
Detailed Explanation
ΔG° = -RT ln K. ln K = -ΔG°/(RT) = -(-5000 J)/(8.314 × 298) = 5000/2477.6 = 2.018. K = e^(2.018) = 7.52 ≈ 7.4.
Hint
Use ΔG° = -RT ln K. Ensure units are consistent (J, not kJ).