Sciencemadness Discussion Board

Synthesis of HBr by photolysis of bromine water

Hexavalent - 27-4-2012 at 11:43

In the thread below, it is very briefly mentioned that under light bromine solutions in whater photolyse into oxygen and hydrobromic acid. Would this be a viable synthesis route, i.e. preparing a bromine solution in water, placing it in a clear glass vessel and exposing it to large, shop lights for a few hours/days to obtain HBr?

Link to the other thread;

http://www.sciencemadness.org/talk/viewthread.php?tid=5572

AndersHoveland - 27-4-2012 at 12:47

"The presence of 10 percent hydrochloric acid prevents all decomposition of chlorine water, even after long exposure to sunshine. Aqueous solutions of pure bromine and iodine water have been exposed to sunlight for a period of fourteen months. It was found that in a dilute solution of bromine water, as much as 57 percent of the total bromide is converted into hydrogen bromide; in a saturated solution, the minimum ammount of decomposition occurs, but increase follows further additions of bromine."

Br2 + 2 H2O + sunlight --> 2 HBr + O2

Note that chlorine will oxidize bromine water to bromic acid.

6 Cl2 + Br2 + 7 H2O --> 2 HBrO3 + 12 HCl (aq)

[Edited on 27-4-2012 by AndersHoveland]

AJKOER - 27-4-2012 at 17:08

Per THE CHEMICAL ELEMENTS AND THEIR COMPOUNDS, by N. V. SIDGWICK, page 1165 ( in our library), to quote:

"Owing to the small heat of formation of hydrogen bromide it is scarcely practical [462] to produce it by the direct combination of the elements, and, its formation is therefore usually supported by some other form of chemical energy, as by combining the bromine with phosphorus or aluminium and treating the product with water: by the action of bromine on an organic substance, such as benzene or naphthalene, when, of course, half the bromine goes to brominate the hydrocarbon: or by treating a salt of HBr with a strong acid (best with phosphoric, as this does not oxidize the HBr produced)."

After reading this, I would suggest:

(1) React Al and Br2 and add H2O to form HBr and Al(OH)3:

2 Al + 3 Br2 → Al2Br6

Al2Br6 + 6 H2O --> 2 Al(OH)3 + 6 HBr

(2) React Br2 + H2O with NaOH and then add H3PO4. I would expect a reduced yield due to the formation of Sodium bromate (NaBrO3).

(3) Per Wikipedia: "The acid may be prepared by several other methods, as well, including reaction of bromine either with phosphorus and water, or with sulfur and water[11]:

2 Br2 + S + 2 H2O → 4 HBr + SO2 "

to which I would also employ sunlight.

(4) Inserting a piece of Aluminum in Bromine water may also help catalyze the reaction (speculation based on the first recommendation) in the presence of sunlight.



[Edited on 28-4-2012 by AJKOER]

Hexavalent - 28-4-2012 at 01:27

Thanks a lot, that cleared up a few points significantly.

The point is, perhaps I should have been more clear, I wish to prepare hydrobromic acid without bromine, and want to attempt a different method to the normal one involving the action of sulfuric acid on the bromide salt.

3)- When the HBr has been made, wouldn't it just react with the aluminium?

As soon as I receive my bromide salt, I will attempt it nonetheless and report back.


PS. I was going to make the bromine solution in a similar way to how iodine is prepared - adding the aqueous salt to hydrochloric acid and hydrogen peroxide under a hood.

AJKOER - 28-4-2012 at 08:27

With respect to the reaction of Al in the presence of Bromine water, my reasoning (speculation) is whether the Br2 is capable of attacking the Aluminum (actually first the Al2O3 protective coating), or the HBr is, or both attack, a common product is aqueous Al2Br6. Upon hydrolysis:

Al2Br6 + 6 H2O --> 2 Al(OH)3 (s) + 6 HBr

This reaction might proceed very slowly, but faster than Br2 and water, hence my catalytic comment.

AndersHoveland - 28-4-2012 at 10:42

Anhydrous AlBr3 supposedly melts at only 97.8 °C, so elemental bromine could probably be refluxed with the aluminum. The AlBr3 that forms on the surface would immediately melt off, allowing the aluminum underneath to react.

Another idea would be the reaction of dry H2S gas with bromine. (unfortunately, H2S cannot reduce elemental iodine to anhydrous hydrogen iodide, but aqeuous hydroiodic acid can be made by this method)
Quote:

Preparation of 57% HI Solution (Hydriodic Acid):
Boiling point: 125.5-126.5 deg C
Density 1.70 gr/ml

In a fume cupboard a 1.5 three neck flask is added a mixture of 480 grams of iodine and 600 mls of distilled water. The centre neck is added a sealed stirring unit which leads almost to the bottom of the flask. The left neck is fitted and sealed with a glass tube which extends almost to the bottom of the flask but does not touch the stirrer. This is connected to the hydrogen sulphide (H2S) generator. The right neck is fitted and sealed with a short glass tube connected with a plastic tube to the bottom of an inverted funnel in a 5% sodium hydroxide solution.

The mixture is vigorously stirred and a stream of hydrogen sulphide (H2S) gas is passed rapidly in the iodine/water mixture as rapidly as it reacts with the iodine. After several hours, 2-3 hours, the iodine disappears and the liquid assumes a yellow colour (sometimes almost colourless) and most of the precipitated sulphur sticks together in the form of a hard lump. This liquid contains a mixture of HI, H2S and S. The liquid is now filtered through a large funnel plugged with glass wool to remove the sulphur S. No need removing the dissolved H2S as this enhances the reductive power. Add a few crystals of iodine to this HI/H2S solution and store at 0-5 deg C.

The sulfur mass can latter be removed by soaking in sodium sulfite solution.

for a chemical explanation of why anhydrous hydrogen iodide cannot form, see: http://www.sciencemadness.org/talk/viewthread.php?tid=17604
Quote: Originally posted by AJKOER  

"Owing to the small heat of formation of hydrogen bromide it is scarcely practical [462] to produce it by the direct combination of the elements,

This is not entirely true. There is a procedure for making anhydrous HBr by slowly passing H2 and Br2 vapor through a copper tube heated to a dull red heat.
http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv2...

[Edited on 28-4-2012 by AndersHoveland]

AJKOER - 28-4-2012 at 19:34

AndersHoveland:

Your comment on disputing the statement "Owing to the small heat of formation of hydrogen bromide it is scarcely practical [462] to produce it by the direct combination of the elements", may not be correct as the synthesis of slowly passing H2 and Br2 vapor through a copper tube heated to a dull red heat, as I understand it, requires Cu (note the parallel to other synthesis employing intermediaries of Al and P). For example, if the chemistry is (as suggested by your source, to quote "7. The copper turnings remove any trace of bromine that may be present by converting it to black cupric bromide. If the level of blackened copper rises, bromine is being carried over" ) the reaction sequence could be:

Cu + Br2 = CuBr2

H2 + CuBr2 = 2 HBr + Cu

where if all the Copper is consumed, Bromine and not HBr is formed in accord with the comment from your source. As such, it may be that this synthesis is not an example of a direct combination of H2 and Br2, and the author's statement remains largely correct.

For support on the different on the Br2 and H2 reaction versus Cl2 and H2, reference: "Reaction of Gas-Phase Bromine Atom with Chemisorbed Hydrogen Atoms on a Silicon(100)-(2´1) surface", by Jongbaik Ree, Kyung Soon Chang, Kyeong Hwan Moon, and Yoo Hang Kim, published in Bull. Korean Chem. Soc. 2001, Vol. 22, No. 8 page 889. To quote:
"However, if the gas-phase atom is bromine, even though it belongs to the same halogen group as chlorine, it doesn’t lead to a highly exothermic reaction but to a slightly endothermic reaction. Thus, in the case of bromine we can expect the reaction mechanism to be quite different from the highly exothermic chlorine or hydrogen reactions." Link: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&a...
-----------------------------------------------------------------------------------------

Also, per Wkipedia on Hydrobromic acid: HBr "can also be prepared by treating bromides with non-oxidising acids like phosphoric or acetic acids." It would be interesting if strong vinegar (like "Spirits of Vinegar") and H2O2 could liberate Bromine from a Bromide.

With respect to your suggestion of treating Bromine water with H2S, it may be easier and safer to just use Sulfur. To quote Wikipedia: "The acid may be prepared by several other methods, as well, including reaction of bromine either with phosphorus and water, or with sulfur and water[11]:

2 Br2 + S + 2 H2O → 4 HBr + SO2 "

Interestingly, this parallels the reported reaction of Cl2 in water with Sulfur, as:

2 Cl2 + 2 H2O <---> 2 HOCl + 2 HCl

S + 2 HOCl + H2O ---> 2 HCl + H2SO3

Or, on net:

2 Cl2 + S + 3 H2O --> 4 HCl + H2SO3

However, I would not expect the reaction to be as rapid as in the case of Chlorine water.

Source: Per Watts' Dictionary of Chemistry Volume 2 Page 16, even dilute solution of HClO can oxidize Sulfur all the way to H2SO4. To quote the relevant section from Watts':

"Reactions.--1. HClOAq acts generally as an oxidiser; it easily parts with 0 while HClAq remains. Thus, As is rapidly oxidised with evolution of light; P, S, Se, Br, I are converted to H3P04Aq, H2S04Aq, &c., even by dilute HClOAq; lower oxides or salts are converted into higher, e.g. SO2Aq to H2SO4Aq, FeO to Fe203, As203Aq to As2O5Aq, FeS04Aq to Fe2(S04)3Aq, Fe2Cl6Aq, and Fe2O3, MnSO4Aq to MnO2; sulphides yield sulphates, c.g. H2SAq gives" H2SO4,Aq and S; "

Link:
http://books.google.com/books/reader?id=ijnPAAAAMAAJ&dq=...


[Edited on 29-4-2012 by AJKOER]

Hexavalent - 30-4-2012 at 14:08

Thanks for all your help guys, really useful. Thanks for the link as well Ajkoer. Is it me, or do the halogens seem to have the dickiest chemistry of them all?:)