Metabolic Rate of a Living Person An average individual (~ 75 kg), consumes the
ID: 485527 • Letter: M
Question
Metabolic Rate of a Living Person An average individual (~ 75 kg), consumes the equivalent to 6, 800 kJ (of food) per day. We would like to have an estimate of his/her metabolic rate. Assuming that all food-derived energy is derived from glucose, and that glucose is metabolized according to the following reaction. This reaction is known to release energy (at a rate of approximately 2800 kJ per mol of glucose). Find the average individual metabolic rate in terms of moles of oxygen consumed per liter (of person, assume that the body is mainly water with a density of 1 kg/L) per Second Explain your analysis, equations, and assumptions.Explanation / Answer
Ans.
#1. Volume of the body = Mass / density
= 75 kg / (1 kg/ L)
= 75 L
Total moles of glucose require to complete daily energy intake =
Daily energy requirement / energy yield of glucose
= 6800 kJ / (2800 kJ/ mol)
= 2.42857 mol
So, a person requires to consume 2.42857 mol glucose per day.
#2. According to the stoichiometry of balanced reaction, the oxidation of 1 mol glucose requires 6 mol O2.
So,
Moles of O2 required to oxidize 2.42857 mol =
(6 O2 / mol glucose) x 2.42857 mol glucose
= 14.57142 mol
That, complete oxidation of 2.42857 mol glucose requires 14.57142 mol O2.
#3. Average metabolic rate = moles of O2 consumed per L body volume per L second.
We have,
O2 consumption =14.57142 mol per 75 L for 24 hours
= (14.57142 mol / 75 L) for 24 hours
= 0.1942857 mol L-1 for 24 hours
= (0.1942857 mol L-1) / 86400 s ; [1 day = 86400 s]
= 2.248 x 10-6 mol L-1s-1
Thus, the metabolic rate in terms of O2 consumption = 2.248 x 10-6 mol L-1s-1