When free or molecular oxygen participates in the respiratory breakdown of organic substrate, it is called aerobic respiration.
C6H12O6 + 6O2 −−−−−−−−−−−−→ 6CO2 + 6H2O + Energy
(686 kcal or 2870 kcal or 38 ATP)
Each glucose molecule contains 686 kcal of energy.
38 ATP x 7.3 kcal = 277.4 kcal
38 ATP x 30.6 kJ = 1162.8 kJ
Thus, in aerobic respiration of each glucose molecule, out of the 686 kcal, 277.4 kcal energy is conserved as usable form. This comes to nearly 40%.
277.4/686 x 100= 40%
As 40% energy is conserved, the efficiency of aerobic respiration is 40%.
C6H12O6 −−−−−−→ 2C2H5OH + 2CO2 + Energy
In anaerobic respiration, 2 ATP molecules are used for breakdown.
2 ATP x 7.3 kcal = 14.6 kcal
2 ATP x 30.6 kJ = 61.2 kJ is conserved in anaerobic respiration of each glucose molecule.
14.6/686 x 100= 2.12%
Thus, energy efficiency of anaerobic respiration is just 2.12%.
This proves that anaerobic respiration is less efficient than aerobic respiration.