Activation Energy and Spontaneity
The combustion of glucose (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O) is a highly exergonic reaction with ΔG° = -2880 kJ/mol. Which of the following best explains why this thermodynamically spontaneous reaction does not proceed at a measurable rate at room temperature?
A
Endothermic nature of the overall reaction
B
High activation energy requiring an initial input of energy
C
Positive entropy change opposing the reaction
D
Equilibrium constant less than 1
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