7. The equation for the oxidation of glucose is C,H,,06 (s) + 6 O2 (g) # 6 H2O (

Question

7. The equation for the oxidation of glucose is C,H,,06 (s) + 6 O2 (g) # 6 H2O (l) + 6 CO2(g) The standard free energy of this combustion reaction, -2880 kJ/mol, is clearly very spontaneous. Why doesn't glucose burst into flame in air? Predict the magnitude and sign of ??' and ?S®. Explain. Much of the -2880 kJ of free energy of glucose oxidation is ultimately trapped by ADP phosphorylation in the mitochondria, but not all. How much is lost, and where? Which steps are most/least efficient? Why is the process not 100% efficient? (note: this is a Fermi question, i.e. an order-of-magnitude estimate of where the energy goes, not arn exact number)

Explanation / Answer

This indeed is an exergonic reaction(means that a reaction is spontaneous) as Gibbs free energy is negative. The glucose doesn't burst into flames because it does not have adequate activation energy. Some exergonic reactions do not occur spontaneously. Instead, the reaction needs an adequate external energy to start. This outside energy is called the ACTIVATION ENERGY of the reaction.

For a spontaneous reaction, ?H should be negative and ?S should be positive (always)

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