The metabolism of glucose, C6H12O6, yields carbon dioxide, CO2(g), and water, H2
ID: 994849 • Letter: T
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