Answer:The efficiency of the aerobic respiration pathway is 39.1%
Explanation:
Efficiency = (Energy output÷Energy input) × 100
Energy output = Amount of energy in 36 ATP molecules = 36 × Amount of energy in 1 ATP molecule = 36×30.5KJ/mol = 1098KJ/mol
Energy input = Amount of potential energy in 1 molecule of glucose burned which is 2805KJ/mol
Efficiency =(1098KJ/mol ÷ 2805KJ/mol) × 100 = 39.1%
Efficiency of the aerobic respiration pathway is 39.1%