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.

Substance	ΔH∘f (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.

Answer 1

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