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 = ?

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

throwdolbeau throwdolbeau · Answer 2 · 2022-04-05T14:48:16+02:00

The heat of reaction for the conversion of 1 mole of glucose into formaldehyde is + 636 KJ.

Conversion of 1 mole of glucose into formaldehyde :

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) ------->6 H₂CO(g) + 6O₂(g) ΔH °comb = + 572.9 KJ/ * 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 * 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 * 6

Adding all equations for calculation of ΔH ° rxn :

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

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

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

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

We will get: C₆H₁₂O₆(s) ---------> 6 H₂CO(g)

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

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