The metabolism of glucose, C6H12O6, yields carbon dioxide, CO2(g), and water, H2
ID: 695526 • 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 71.0 % efficiency. Use the data below to answer questions about the metabolism of glucose.
Part A
Calculate the mass of glucose metabolized by a 75.9 kg person in climbing a mountain with an elevation gain of 1610 m . Assume that the work performed in the climb is four times that required to simply lift 75.9 kg by 1610 m .
Express your answer to three significant figures and include the appropriate units.
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 71.0 % 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
Part A
Calculate the mass of glucose metabolized by a 75.9 kg person in climbing a mountain with an elevation gain of 1610 m . Assume that the work performed in the climb is four times that required to simply lift 75.9 kg by 1610 m .
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
The consumption of glucose is
C6H12O6+ 6O2----->6CO2(g)+6H2O(l)
enthalpy change = sum of enthalpy of products- sum of enthalpy of reactants
= 6* enthalpy of CO2+6* enthalpy of H2O(l)- { 1* enthalpy of C6H12O6+6* enthalpy of O2}
=6*(-393.5)+6*(-285.8)- {1*(-1273.3)+6*0}
where 6,6,1 and 6 are coefficients of CO2 (g), H2O(l), C6H12O6 and O2 respectively.
=-2802.5 KJ
work produced = -2802.5*0.4= -1121 Kj this is the energy produced per mole of glucose,
potential energy= mgh
potential energy= 75.9*9.8*1610 kg.m/sec2*m =1183350 Joules = 1183.35 KJ
hence moles of glucose required = 1183.35/1121= 1.06 moles
mass of glucose required = moles* molar mass of glucose = 1.06*180=190.8 gm
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