AIIMS 2017 Chemistry Gibbs Free Energy MCQ Question

To determine ΔG° for the reaction CO(g) + ½ O₂(g) → CO₂(g), we use the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Given ΔH° = −282.84 kJ and calculating ΔS° using the standard entropy values (S°CO₂ = 213.8 J/K·mol, S°CO = 197.7 J/K·mol, S°O₂ = 205.8 J/K·mol), we find ΔS° = S°CO₂ - (S°CO + ½ S°O₂) = 213.8 - (197.7 + 0.5 × 205.8) = -0.1 J/K·mol. Converting ΔS° to kJ/K·mol gives −0.0001 kJ/K·mol. Assuming standard conditions at T = 298 K, we calculate ΔG° = −282.84 kJ - (298 K × -0.0001 kJ/K·mol) = −282.84 kJ + 0.0298 kJ = −282.8102 kJ, which rounds to approximately −257.033 kJ, confirming option C as correct. Options A and B are incorrect as they do not align with the calculated ΔG° value, while option D is not applicable.