The metabolism of glucose, C6H12O6, yields carbon dioxide, CO2(g), and water, H2

Question

The metabolism of glucose, C6H12O6, yields carbon dioxide, CO2(g), and water, H2O(l), as products. Energy released in this metabolic process is converted to useful work, w, with about 68.0 % efficiency. Use the data below to answer questions about the metabolism of glucose.

Substance Hf (kJ/mol)

CO2(g) 393.5

C6H12O6(s) 1273.3

H2O(l) 285.8

O2(g) 0

Calculate the mass of glucose metabolized by a 74.6 kg person in climbing a mountain with an elevation gain of 1590 m . Assume that the work performed in the climb is four times that required to simply lift 74.6 kg by 1590 m .

Explanation / Answer

first, let's write the reaction of combustion of glucose:

C6H12O6 + 6O2 = 6CO2 + 6H2O

dHrxn = (-6*285.8 - 6*393.5) - (-1273.3) = 2802.5 kJ/mol

W = work = mgh
W = 74.6 * 9.8 * 1590 * 4 = 4.65x106 J or simply 4.65x103 kJ

EDITED: the Work is with 68% efficiency so:
W = 4.65x103 * 0.68 = 3.162 kJ

Now let's calculate the moles:
moles glucose = 3.162x103 / 2802.5 = 1.1283 moles

m = mol * MW
m = 1.1283 * (180.16) = 203.27 g

Hope this helps

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