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AP Chemistry/Unit 9: Applications of Thermodynamics
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Activation Energy and Spontaneity
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Which property would BEST explain why the combustion of glucose (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, ΔG° = -2880 kJ/mol) doesn’t occur spontaneously at room temperature despite being highly exergonic?

A

Positive entropy change opposing the reaction

B

High activation energy requiring an initial input of energy

C

Equilibrium constant less than 1

D

Endothermic nature of the overall reaction

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