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.

A) Write a balanced chemical reaction for the combustion of glucose.

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

C) 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?

D) 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

Solution A

The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) -->6CO2(g) + 6H2O(g).

Solution B

C + O2 ------> CO2 H= -394 kJ

H2 + 1/2 O2 -------> H2O H= -286kJ

6C + 6H2 + 3O2 -------> C6H12O6 H= -1170kJ

Now, to calculate the enthalpy of combustion, we have:

H = 6 * Hof[CO2(g)] + 6 * Hof[H2O(g)] Hof[C6H12O6(s)]

H = 6 * (393.5 kJ/mol) + 6 * (285.8kJ/mol) [1273.3+6(0kJmol)] = 2802.5kJ/mol

Solution C

2802.5 kJ of energy will be released by the combustion of 1 mole (180 gm) of glucose.

Weight of glucose required for 2000kJ = 180*2000/2802.5 gm = 128.45 gm

Solution D

The Hess's law facilitates us to calculate the metabolic energy despite multiple steps in the process.

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