The chemical equation for this is: <span>
<span>C6H12O6 +
O2 --> H2O + CO2 </span></span>
By balancing the reaction, we get:<span>
<span>C6H12O6 + 6O2 --> 6H2O + 6CO2</span></span>
By doing an overall energy balance, we can see that the work
expansion is simply equivalent to the heat released.
To get the heat released, we use enthalpies of formations for
the different substances: <span>
HF(glucose) = -1250 kJ/kmol
HF(liquid water) = -285.83 kJ/mol
HF(water vapour) = -241.82 kJ/mol
HF(carbon dioxide) = -393.52 kJ/mol
<span>HF(oxygen) = 0
For a) liquid water</span></span>
<span>
H(Total) = (HF glucose) – 6 (HF liquid water) – 6 (HF carbon
dioxide) </span>
H(Total) = -2826.1 kJ/mol = Work <span>
<span>For b) water vapour</span></span>
<span>
H(Total) = - (HFglucose) + 6 (HFwater vapour) + 6 (HFcarbon
dioxide) </span>
H(Total) = -2562.04 kJ/mol. <span>
However this is still in per mole basis, so calculate number
of moles of glucose supplied:
molar mass = 180.16 g/mol
moles glucose = 1g / 180.16 g/mol = .000555 mol
a) Work = -2826.1 kJ/mol * 0.000555 = -1.568 kJ
<span>b) Work = -2562.04 kJ/mol * 0.000555 = -1.42 kJ </span></span>