Values of Hfo from Standard Heats of Combustion
To measure directly the heat of formation of sucrose, C12H22O11, you would have to carry out the following reaction:
 
12 C(s) + 11 H2(g) + 5½ O2(g) ------> C12H22O11

 
But no one has ever been able to figure out how to make this reaction occur directly under any conditions, so there is no direct way to measure the Hfo of sucrose. How, then, can we obtain values of Hfo for compounds such as sucrose?
 
If the compound in question can be burned - which is usually far easier to do than make it from its elements - then we have a source of energy data that we can use to calculate its Hfo. This is because the products of combustion are nearly always compounds whose values of Hfo are known or can be measured by direct means. The combustion of sucrose in an atmosphere of pure oxygen proceeds by the following equation:
 
C12H22O11(s) + 12 O2(g) -----> 12 CO2(g) + 11 H2O(l)

 
If the standard enthalpy change for this reaction can be measured, and if we can look up the values of Hfo for three of the four chemicals in the equation, then we can use the Hess law equation to find the Hfo of the remaining substance, sucrose.
 
The above equation is a combustion equation and your data tables have Hco values (standard heat of combustion). Putting these into a Hess's Law equation you should get:
 
Hco = [(12)CO2(g) + (11)H2O(l)] - [C12H22O11(s) + (12)O2(g)]

 
All values are in kJ/moles.

          -5639.7 = [(12)-393.5 + (11)-285.8] - [C12H22O11(s) + (12)0]
          -5639.7 = -4722 -3143.8 -[C12H22O11]
            2226.1 = -[C12H22O11]
 

Therefore the Hfo of sucrose, C12H22O11, is -2226.1 kJ/mole
 
Sample problem
One of the "building blocks" for proteins such as those in muscles and sinews is an amino acid called glycine, C2H5NO2. The equation for its combustion is
 
4 C2H5NO2(s) + 9 O2(g) ---> 8 CO2(g) + 10 H2O(l) + 2 N2(g)
The value of Hco for glycine is -973.49 kJ/mole. Using this information and the values of Hfo calculate the Hfo for glycine.
 
Solution
For this problem, Hess's law equation becomes
 
Ho = [(8)CO2(g) + (10)H2O(l) + (2)N2(g)]-[(4)C2H5NO2(s) + (9)O2(g)]
 
No, I didn't forget the Hco. The first term Ho, is obtained from the standard heat of combustion of glycine. Since the chemical equation for this reaction is for the combustion of four moles of glycine, we have to multiple Hocombustion by four.
 
Ho = 4 mol x -973.49 kJ/mol = -3894.0 kJ
 
Now we can substitute into Hess's law equation the correct values.

     -3894.0 kJ = [(8)-393.5 + (10)-285.8 + (2)0] - [(4)C2H5NO2 + (9)0]
         -3894.0 = -3148.0 -2858 -[(4)C2H5NO2]
          2112.0 = -[(4)C2H5NO2]
 

Therefore by rearranging we get

                 Hfo for C2H5NO2 = -2112.0 kJ/mole = -528.0 kJ/mole
                                                      4 moles
 

Thus, the standard heat of formation of glycine is -528.0 kJ/mol, and we have seen how we can determine this quantity without making glycine directly from its elements.
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