Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings, and plastics. Cellulose consists of long chains of glucose molecules (C6H12O6), so for the purposes of modeling the reaction we can consider the conversion of glucose to formaldehyde (H2CO).
1. Calculate the heat of reaction for the conversion of 1 mole of glucose into formaldehyde, given the following thermochemical data:
H2CO(g) + O2(g) > CO2(g) + H2O(g) ΔH °comb = - 572.9 KJ/mol
6 C(s) + 6 H2(g) + 3 O2(g) > C6H12O6(s) ΔH ° f = - 1274.4 KJ/mol
C(s) + O2(g) > CO2(g) ΔH ° f = - 393.5 KJ/mol
H2(g) + 1 / 2 O2(g) > H2O(g) ΔH ° f = - 285.8 KJ/mol
C6H12O6(s) > 6 H2CO(g) ΔH ° rxn = ?

+ 636 KJ

Explanation:

We want to arrive to the equation

C6H12O6(s) > 6 H2CO(g) ΔH ° rxn = ?

by manipulating algebraically the first four  given equations.

We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6

6 CO2(g) + 6 H2O(g)  >6H2CO(g) + 6O2(g)  ΔH °comb = + 572.9 KJ/x 6

Now we want C6H12O6(s) as a reactant and it  is a product in the second one, therefore lets reverse it

C6H12O6(s)  > 6 C(s) + 6 H2(g) + 3 O2(g)   ΔH ° f = + 1274.4 KJ/mol

Now if take the third equation and multiply it by six we will cancel the C(s) with the above equation

6 C(s) + 6O2(g) > 6 CO2(g) ΔH ° f = - 393.5 KJ/mol x 6

Finally by multiplying the last equation by 6 and adding all the equations we will arrive at our desired one

6 H2(g) + 3 O2(g) > 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6

then lets add them to get ΔH ° rxn:

  6 CO2(g) + 6 H2O(g)  >6H2CO(g) + 6O2(g)  ΔH °comb = + 3437.4 KJ

+ C6H12O6(s)  > 6 C(s) + 6 H2(g) + 3 O2(g)      ΔH ° f = + 1274.4 KJ

+ 6C(s) + 6O2(g) > 6 CO2(g)                            ΔH ° f = - 2361.0 KJ

+6 H2(g) + 3 O2(g) > 6H2O(g)                      ΔHº f  = - 1714.8 KJ

C6H12O6(s) > 6 H2CO(g)  

ΔH ° rxn =  3437.4 + 1274.4 - 2361.0 - 1714.8 =  636 KJ

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