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Reaction mechanisms in the carbothermic reduction of phosphoric acid


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The carbothermic reduction of phosphoric acid is thought to be represented by the reaction:

4H3PO4 + 16C → 6H2 + 16CO + P4

The equilibrium equations for orthophosphoric acid are:

H3PO4 + H2O ↔ H3O+ + H2PO4-

H2PO4- + H2O ↔ H3O+ + HPO42-

HPO42- + H2O ↔ H3O+ + PO43-

Condensation reactions can join phosphoric acid molecules:

2H3PO4 → H4P2O7 + H2O

H4P2O7 + H3PO4 → H5P3O10 + H2O

Condensation between two -OH units of the same molecule can create a cyclic molecule:

H5P3O10 → H3P3O9 + H2O

Various_Phosphoric_Acids.PNG

The chemical structure of activated carbon is closely approximated by the structure of graphite:

800px-Graphite_unit_cell.gif

The thermodynamics of H3PO4 vaporization was studied by Brown and White in "Vapor pressure of phosphoric acids" (1952). As the acid is heated P4O10/mass% moves towards the azeotropic equilibrium with a composition of approximately 92% P4O10. In the Huhti and Gartaganis paper "The composition of the strong phosphoric acids" (1956) the percent of "Hypoly-" phosphoric acids at 86.26 wt. % P2O5 was found to be 66.03%, and in general there was a monotonic increasing relationship between these two percentages. From this we can surmise that the carbothermic reduction of phosphoric acid involves reductions of hypoly- and -meta phosphoric acid molecules, but what are the precise mechanisms?

Mechanisms which immediately came to my mind included homolytic cleavage + radical mechanisms, and dehydration, but it is difficult to visualize how this would play out in 3-d space at the molecular level, especially since it probably requires a precise accounting of the activity of sigma and pi bonds, and the unpaired elections left over after a homolytic cleavage event. I am continuing to hack away at this problem slowly with drawings to increase my visual memory while I try to visualize what is happening at the molecular level, but I am asking for your thoughts and help in solving this complicated problem.

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1 hour ago, John Cuthber said:

I think the best guess is attack by P4O10 vapour on the graphite.

Homolytic thermolysis to give smaller molecules- notable P4O3 -is probably a feature too.
But I doubt anyone really knows.

Do you think there would be much of the anhydride in the azeotropic mixture? AFAIK the only way to prepare phosphorus pentoxide is by burning elemental phosphorus, but that may only be due to the difficulty of extracting the anhydride from the azeotropic mixture of very hot vaporized phosphoric acid, if it is even present there at all.

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