Complete oxidation of glucose by burning in a calorimeter gives a $\Delta$G°´ of -2870 kJ/mol. If we assume this to be the maximum energy that can be extracted from glucose, what is the percent of energy that is captured in a cell given that about 32 ATP can be made from a single molecule of glucose? $\Delta$G°´ for ATP hydrolysis is -30.5 kJ/mol while $\Delta$G for ATP hydrolysis in a cell is about -51 kJ/mol.

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The Answer of Complete oxidation of glucose by burning in a calorimeter gives a $\Delta$G°´ of -2870 kJ/mol. If we assume this to be the maximum energy that can be extracted from glucose, what is the percent of energy that is captured in a cell given that about 32 ATP can be made from a single molecule of glucose? $\Delta$G°´ for ATP hydrolysis is -30.5 kJ/mol while $\Delta$G for ATP hydrolysis in a cell is about -51 kJ/mol.

Complete Oxidation of Glucose by Burning in a Calorimeter Gives Δ\DeltaΔ

Complete Oxidation of Glucose by Burning in a Calorimeter Gives $\Delta$