Glucose (C 6 H 12 O 6 ) is the primary source of metabolic energy for vertebrate

Question

Glucose (C6H12O6) is the primary source of metabolic energy for vertebrates.

Write a balanced chemical reaction for the combustion of glucose.

Using standard enthalpies of reaction, calculate the enthalpy of reaction for the combustion of glucose.

A normal human body requires approximately 2000 kJ of energy to sustain itself. How many grams of glucose must be consumed to provide 2000 kJ of energy?

The metabolism of glucose in vertebrate systems requires many steps, which when added together result in the reaction from part (a). Explain why it is possible to calculate the metabolic energy even though there is more than one step in the process.

Explanation / Answer

Combustion of glucose

reaction,

C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)

Enthalpy of reaction using standard enthalpy of formation values,

dHof(rxn) = dHof(products) - dHof(reactants)

                = (6 x -393.5 + 6 x -285.8) - (-1275 + 6 x 0)

                = -2801 kJ/mol

A normal human body requires 2000 kJ of heat

amount of heat produced by combustion of 1 mole (180.16 g) of glucose = 2801 kJ

so to produce 2000 kJ heat, amount of glucose needed = 2000 x 180.16/2801 = 128.64 g

The metabolism in the body requires several steps, (a) however it is possible to calculate the metabolic energy even thought there is more than one step as the final product of metabolism remains the same as seen above and thus the overall reaction remains unchanged even after several steps of metabolism.

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