Standard Enthalpy of Formation?

In the calorimeter, the reaction that took place was:

C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(l)

From the temperature change of the calorimeter and the heat capacity of the calorimeter, you can calculate the heat absorbed:

q = (7.793 K X 641 J/K) = 4.995X10^3 J

So, the combustion reaction released -4.995X10^3 J

The moles of glucose burned in that reaction was:

0.3212 g / 180.156 g/mol = 1.783X10^-3 mol glucose

So, the enthalpy change for the combustion of glucose is -4.995X10^3 / 1.783X10^-3 mol = -2.80X10^6 J/mol / -2.80X10^3 kJ/mol

Now, you know that for any chemical equation,

Delt Hrxn = sum of Delta Hf products - sum of Delta Hf reactants. So:

-2800 kJ/mol = [6(-393.5) + 6(-286 kJ/mol)] - Delta Hf glucose

Delta Hf glucose = -1277 kJ/mol