Olympian Carbon Footprint

Chemistry Level 3

In the recently completed Olympic Games in Rio, the winning time in the women's marathon was 2 hours, 24 minutes and 4 seconds. (For the sake of this problem, 4 seconds has been subtracted from the real record.) Suppose the winning runner consumed 2.8 liters of O X 2 \ce{O2} per minute (average) during the race (2.8 liters/min at standard temperature and pressure). Assume also that 100 percent of her energy was supplied by the oxidation of glucose.

C X 6 H X 12 O X 6 + O X 2 → C O X 2 + H X 2 O \ce{C6H_{12}O6 + O2 -> CO2 + H2O}

Estimate her carbon footprint (in moles) for this race in moles of C O X 2 \ce{CO2} .

Details and assumptions :

  • Metabolism is much more complex than this, with glycolysis, Krebs cycle and electron transport chain plus additional energy from the oxidation of fatty acids, but just assume simple oxidation of glucose gives your estimate.

  • You are given oxygen consumption, not inspired or expired air volume, which would be much greater.

  • Oxygen consumption is stated relative to STP conditions.


The answer is 18.

Matt Owings by Matt Owings , 69, USA

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1 solution

Matt Owings
Aug 27, 2016

We are given her time, 2 2 hours 24 24 minutes, or 144 144 minutes. We are also given that her average rate of oxygen consumption during the race is 2.8 2.8 liters per minute. Therefore the total volume of O X 2 \ce{O2} she consumes is 144 × 2.8 = 403.2 144 \times 2.8 = 403.2 Liters.

Physiological gas exchange measurements are stated relative to STP (her actual gas volumes on the course would differ...higher...but 403.2 403.2 liters is the value corrected to STP).

We know that one mole of gas at STP has a volume of 22.4 22.4 Liters. So her oxygen consumption is 403.2 22.4 = 18 mol \dfrac{403.2}{22.4} = \SI{18}{\mole} of O X 2 \ce{O2} .

Balancing our equation

C X 6 H X 12 O X 6 + O X 2 → C O X 2 + H X 2 O \ce{C6H12O6 + O2 -> CO2 + H2O}

This gives us

C X 6 H X 12 O X 6 + 6 O X 2 → 6 C O X 2 + 6 H X 2 O \ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}

6 6 moles of O X 2 \ce{O2} consumed yield 6 6 moles C O X 2 \ce{CO2} , ⟹ \implies 18 18 moles O X 2 \ce{O2} consumed yield 18 18 moles C O X 2 \ce{CO2} .

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