Applications of the General Mole Balance Metabolic Rate of a Living Person An av

Question

Applications of the General Mole Balance Metabolic Rate of a Living Person An average individual (~75 kg), consumes the equivalent to approximately 6,800 KJ (of food) per day. We would like to estimate of his/her metabolic rate. Assuming that all food-derived energy is derived from that glucose is metabolized according to the following reaction. This reaction is known to release energy (at a rate of approximate y 2800 kJ per mol of glucose). Find the average individual metabolic rate in terms of moles of oxygen consumed per litter (of person, assume that the body is mainly water with a density of 1 kg/L) per second. Explain your analysis, equations, and assumptions. Note: M_oxygen= 32 kg/kmol, M_glucose= 180 kg/kmol

Explanation / Answer

volume of average person = 75 kg/ 1 kg/L = 75 L

6800 kJ energy is produce / day

therefore moles of glucose used per day = 6800 kJ/ 2800 kJ/ mol = 2.43 mol

Hence average moles of oxygen consumed per day = 2.43 x 6 =14.57 mols ( from the balanced equation)

Therefore average metabolic rate = 14.57 mols/ 75 L = 0.194 mol/ L

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