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Author: Subject: potassium bisulfate + Hydrogen peroxide
smartgene
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[*] posted on 3-8-2016 at 11:28
potassium bisulfate + Hydrogen peroxide


I was wondering reacting potassium bisulfate with pure hydrogen peroxide or sodium percarbonate or may urea peroxide will produce oxone

KHSO4 + H2O2 (conc.) = KHSO3(O2) + H2O (0°C).
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Oscilllator
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[*] posted on 3-8-2016 at 20:48


It looks like it might exist in equilibrium to some extent, but how would you remove the oxone to drive the reaction to completion?
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Deathunter88
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[*] posted on 3-8-2016 at 21:08


Also where would you get pure hydrogen peroxide? Pure hydrogen peroxide is very unstable and can decompose violently.
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smartgene
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[*] posted on 4-8-2016 at 04:12


remove the water and I read that sodium percarbonate is a pure source of hydrogen peroxide
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PHILOU Zrealone
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[*] posted on 4-8-2016 at 07:29


Quote: Originally posted by smartgene  
remove the water and I read that sodium percarbonate is a pure source of hydrogen peroxide

Very smart!
Just like 50% H2O2 is a pure source of H2O2 if you remove the 50% water from it.

You have to remove the carbonate too...

Urea peroxyde complex is quite interesting stable form of H2O2...H2N-CO-NH2.H2O2 is roughly equivalent to 36% H2O2 but solid and with urea instead of water...




PH Z (PHILOU Zrealone)

"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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AJKOER
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[*] posted on 4-8-2016 at 08:12


First, create a bsulfite:

SO2 + H2O ⇔ H+ + HSO3- k =1.3x10-2

Note: HSO3- ⇔ H+ + SO3(2-) k =6.4x10-8

Or, alternately, by adding a metasulfite (like K2S2O5) to water:

S2O5(2−) + H2O ⇌ 2 HSO3−

Now, one needs to create a hydroxyl radical, OH• . One convenient method is by the action of strong sunlight on an aqueous solution of N2O (Laughing gas), or the action of UV on H2O2. But as H2O2 + HSO3- forms HSO4- unless pH is too high (see http://pubs.acs.org/doi/abs/10.1021/j100587a005 ), the use of hydrogen peroxide as a hydroxyl radical source likely requires adding a base.

Next, the action of the newly created hydroxyl radical on hydrogen sulfite (or the sulfite) ion:

OH• + HSO3- → SO3•- + H2O (A11) 2.7xE09
0H• + SO3(2-) → SO3•- + OH- (A12) 4.6xE09

Then, the comparatively slow action of oxygen on the sulfite radical anion (likely requiring aeration with an air pump):

SO3•- + O2 → SO5•- (A17) 1.3xE03

Finally, the formation of the peroxymonosulfate:

SO5•- + HSO3- → HSO5- + SO3•- (A135)

SO5•- + SO3(2-) + H2O → HSO5- + SO3•- + OH-

[Edit] However, this may be only a transient presence as other reactions include:

HSO5- + HSO3- + H+ → 2 SO4(2-) + 3 H+

A better route, that avoids adding acidic HSO3-, add HO2• via:

H2O2 + OH• = HO2• + H2O

SO5•- + HO2• → HSO5- + O2

Source: See "Modeling study of strong acids formation and partitioning in a polluted cloud during wintertime - UNIV-TLSE3 - Hal", Table 2, Table 3 and Table 8 at https://www.google.com/url?sa=t&source=web&rct=j&...

Note: Oxone® active ingredient is potassium peroxymonosulfate, KHSO5.

[Edited on 4-8-2016 by AJKOER]

[Edited on 5-8-2016 by AJKOER]
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[*] posted on 4-8-2016 at 11:50


" One convenient method is by the action of strong sunlight on an aqueous solution of N2O (Laughing gas), "
LOL
N2O doesn't absorb the accessible bit of the UV range.

But even better is the idea that you can make oxone in the presence of bisulphite in the bulk phase.
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AJKOER
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[*] posted on 4-8-2016 at 12:31


Actually, my comment relating to the use of aqueous N2O as a source of hydoxyl radicals was sunlight, not UV. Solar light is spread over a wider range, see chart at http://naturalfrequency.com/wiki/solar-radiation

This includes the values needed for aqueous N2O, see https://www.jstor.org/stable/2415174?seq=1#page_scan_tab_con...

It is long known that radiation also produces the hydroxyl radical from aqueous N2O and my experience suggests pulse radiation from a microwave works as well. See http://pubs.acs.org/doi/abs/10.1021/j100453a020?journalCode=...

Interesting, I do recall less favorable comments in a book relating to the action of light on H2O2 to produce hydroxyl radicals. See, for example, https://www.google.com/url?sa=t&source=web&rct=j&...

As whether a stable presence of HSO5- can be maintained, I agree this may be problematic. I have edited my comments and introduce a possible better approach.

My interest concerns only obtaining an active presence of HSO5-, and not in isolating the salt.

[Edited on 5-8-2016 by AJKOER]
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[*] posted on 5-8-2016 at 15:35


Interestingly, just found a partial confirmation on my radical path that notes a 30% presence on the HSO5-, to quote from
"chemical kinetics of intermediates in the autooxidation of so2" - Argonne National Laboratory, Link: https://www.google.com/url?sa=t&source=web&rct=j&... :

"A product with the properties of peroxymonosulfate has been observed with a yield up to 30% upon bubbling oxygen through a solution of sodium sulfite (13). This compound can undergo many possible subsequent reactions. "
(13). Deuvyst, E. A.; Ettel. V. A.; Mosolu, M. A., CHEMTECH, 1979, 426.

I suspect the existing presence of transition metals ions (Mn, Fe, Co,..) provided a path to the sulfite radical anion via Fenton-like reactions or speciation paths, for example:

pH 2.5 to 4: FeOH(2+) + HSO3- = HOFeSO3H(+)
HOFeSO3H(+) → FeOH(+) + HSO3•

pH 4 to 6: FeOH(2+) + HSO3- = (HO)2FeSO3H
(HO)2FeSO3H → Fe(OH)2 + HSO3•

Reference, see, Millero et al. , "Reduction of Fe(lll) with Sulfite", J. of Geophysical Research, Vol 100, #D4, p. 7235-7244.

[Edited on 6-8-2016 by AJKOER]
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[*] posted on 28-9-2016 at 09:02


Quote: Originally posted by AJKOER  
Interestingly, just found a partial confirmation on my radical path that notes a 30% presence on the HSO5-, to quote from
"chemical kinetics of intermediates in the autooxidation of so2" - Argonne National Laboratory, Link: https://www.google.com/url?sa=t&source=web&rct=j&... :

"A product with the properties of peroxymonosulfate has been observed with a yield up to 30% upon bubbling oxygen through a solution of sodium sulfite (13). This compound can undergo many possible subsequent reactions. "
(13). Deuvyst, E. A.; Ettel. V. A.; Mosolu, M. A., CHEMTECH, 1979, 426.

I suspect the existing presence of transition metals ions (Mn, Fe, Co,..) provided a path to the sulfite radical anion via Fenton-like reactions or speciation paths, for example:

pH 2.5 to 4: FeOH(2+) + HSO3- = HOFeSO3H(+)
HOFeSO3H(+) → FeOH(+) + HSO3•

pH 4 to 6: FeOH(2+) + HSO3- = (HO)2FeSO3H
(HO)2FeSO3H → Fe(OH)2 + HSO3•

Reference, see, Millero et al. , "Reduction of Fe(lll) with Sulfite", J. of Geophysical Research, Vol 100, #D4, p. 7235-7244.

[Edited on 6-8-2016 by AJKOER]


My take on a summary of likely reactions:

H2O + SO3(2-) ⇔ OH- + HSO3-

Fe(3+)/Cu(2+) + HSO3- → Fe(2+)/Cu(+) + HSO3• (See speciation detail for iron above)

HSO3•- + O2 + M → HSO5•- + M (reference: equation R2b and also M4 at https://www.google.com/url?sa=t&source=web&rct=j&... )

Fe(+2)/Cu(+) + O2 → Fe(+3)/Cu(+) + O2•-

The above reaction being so called metal autoxidation, see, for example, "Generation of .OH initiated by interaction of Fe2+ and Cu+ with dioxygen; comparison with the Fenton chemistry", by Norbert K. Urbañski and Andrzej Berêsewicz, link: https://www.ncbi.nlm.nih.gov/pubmed/11996118 ) . And, finally:

HSO5• + O2•- → HSO5- + O2

which is based on the cited reaction above SO5•- + HO2• → HSO5- + O2 along with HO2• = O2•- + H+ (pKa = 4.8) and assuming HSO5• = SO5•- + H+ .

Unfortunately, there are other reactions which can consume the HSO5-, for example:

Fe(2+)/Cu(+) + HSO5- → Fe(3+)/Cu(2+) + OH• + SO4(2-)

which is not necessarily a complete loss of the HSO5- as there are reaction paths employing the created hydroxyl radical (detailed previously above) which can provide some recovery.

[Edited on 28-9-2016 by AJKOER]
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