Calculate the work of expansion accompanying the complete combustion of 1.0 g of glucose to carbon dioxide and (a) liquid water, (b) water vapor at 20oc when the external pressure is 1 atm.

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 

h let me think

it is stable as it is in group 0 which have full outer shells of electrons