The chemical equation for this is:
C6H12O6 +
O2 --> H2O + CO2
By balancing the reaction, we get:
C6H12O6 + 6O2 --> 6H2O + 6CO2
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:
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
HF(oxygen) = 0
For a) liquid water
H(Total) = (HF glucose) – 6 (HF liquid water) – 6 (HF carbon
dioxide)
H(Total) = -2826.1 kJ/mol = Work
For b) water vapour
H(Total) = - (HFglucose) + 6 (HFwater vapour) + 6 (HFcarbon
dioxide)
H(Total) = -2562.04 kJ/mol.
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
b) Work = -2562.04 kJ/mol * 0.000555 = -1.42 kJ
