Sciencemadness Discussion Board

ClBr from chlorine replacement?

Gooferking Science - 31-7-2013 at 11:56

I posted a video on youtube about making bromine by this method.
NaBr + HCl --> HBr + NaCl
HBr + Ca(ClO) --> CaCl2 + H2O + Br2

Is it possible that some ClBr will be produced during the time when the chlorine takes the bromine's spot in NaBr? I don't think it is but someone asked me if it was possible in the comments and wanted to be for sure.

kristofvagyok - 31-7-2013 at 12:50

Quote: Originally posted by Gooferking Science  
I posted a video on youtube about making bromine by this method.
NaBr + HCl --> HBr + NaCl
HBr + Ca(ClO) --> CaCl2 + H2O + Br2


Nope.

HBr is a stronger acid than HCl so you just simply dissolved the NaBr in some hydrochloric acid, only a traces of HBr is generated in that solution.
Why? Ka of HBr is 2.2 x 10^10, while the HCl only has a Ka 1.2 x 10^8, it means that HBr is 200 times stronger acid than HCl.

So how the bromine is generated?
Chlorine is produced from the hydrochloric acid and the hypochorite and this chlorine, what is stronger oxidizer than bromine will "oxidize out" the bromine from it's salt.
Cl2 + 2NaBr = 2 NaCl and Br2.

This Br2 is NOT pure, so please do not write on youtube that you have produced high purity bromine.

ClBr is also generated, so if you want to make ClBr free, relative pure bromine, than use a high excess of NaBr compared to the generated chlorine.

Gooferking Science - 31-7-2013 at 14:14

Well I distilled it so it is a little higher purity.
But It is as pure as I need it.
I also dried it with sulfuric acid to rid it of any extra water. It is in ampoules.
[Edited on 31-7-2013 by Gooferking Science]

[Edited on 31-7-2013 by Gooferking Science]

Oscilllator - 31-7-2013 at 20:44

kristofvagyok couldn't you also remove the ClBr (as well as any Cl) by refluxing the bromine over a bromide salt?

ClBr + NaBr -> NaCl + Br2

AndersHoveland - 31-7-2013 at 21:54

Quote: Originally posted by Oscilllator  
ClBr + NaBr -> NaCl + Br2

be careful with that reaction, in dilute solutions bromate can form

yuno - 17-8-2013 at 13:28

Can you explain?
What do you propose for removing ClBr from elemental bromine?

violet sin - 17-8-2013 at 14:33

I think oscilllator meant the dried bromine w/ Cl impurities refluxed over dry bromide salt. Cl portion replaces bound Br and lock NaCl. Not in solution

Oscilllator - 17-8-2013 at 15:00

Correct violet, that is what I meant. I think woelen suggested it first, in another thread.

AJKOER - 20-8-2013 at 08:24


As a reference, see "Equilibrium and kinetics of bromine chloride hydrolysis" by Liu Q, Margerum DW, Department of Chemistry, Purdue University, West Lafayette, Indiana 47907-1393, USA. To quote:

"Abstract
Aqueous-phase halogen reactions play an important role in tropospheric ozone depletion that is observed during Arctic sunrise where bromine chloride is a key intermediate. The temperature dependencies of BrCl(aq) equilibration with BrCl2-, HOBr(aq), Br2(aq), Cl2(aq), HOCl(aq), Br-, and other species (Br3-, Br2Cl-, Cl3-, OBr-, and OCI-) are determined as a function of Cl- concentration from pH 0 to pH 7. Values for K1 (=[BrCl2-]/([BrCl(aq)][Cl-])) at mu = 1.0 M are 3.8 M(-1) at 25.0 degrees C, 4.7 M(-1) at 10.0 degrees C, and 5.5 M(-1) at 0.0 degrees C, with deltaH1 degrees = -9.9 kJ mol(-1) and deltaS1 degrees = -22 J K(-1) mol(-1). BrCl(aq) hydrolysis equilibria have little or no temperature dependence with Kh1 (=[HOBr(aq)][Cl-][H+]/[BrCl(aq)]) = 1.3 x 10(-4) M2 from 25.0 to 5.0 degrees C, mu = 1.0 M. When conditions are adjusted to give a rapid partial hydrolysis of BrCl in equilibrium with HOBr and Cl- at p[H+] 4.31, a relatively slow reaction (kobsd = 2.4 s(-1)) to form HOCl and Br- is observed. This takes place via BrCl reaction with Cl- to form Cl2, which hydrolyzes in the rate-determining step to give HOCl. On the other hand, the rate of complete BrCl hydrolysis to form HOBr and Cl- at p[H+] 6.4 is extremely rapid with a first-order rate constant of 3.0 x 10(6) s(-1) at 25.0 degrees C. The reverse reaction between HOBr, Cl-, and H+ has a rate constant of 2.3 x 10(10) M(-2) s(-1), so that in seawater, where [Cl-]/[Br-] = 700, the formation of BrCl is much faster than the formation of Br2 from HOBr, Br-, and H+. Rapid formation of BrCl(aq) and its subsequent reaction with Br- is a viable pathway to give Br2(aq). Photolysis of Br2(g) is believed to initiate the reactions associated with ozone depletion."

Link: http://www.ncbi.nlm.nih.gov/pubmed/11347924

Now, with the formation of HOBr and the right conditions (concentration, temperature,..) a disproportionation creating some bromate could follow:

3 HOBr --> 2 HBr + HBrO3


[Edited on 20-8-2013 by AJKOER]

woelen - 20-8-2013 at 23:07

In the context of removing chlorine or BrCl from bromine by reacting with NaBr, the formation of bromate hardly is relevant. I think that you can state that the loss of bromine to bromate will be less than 0.1%, in other words, you can neglect that reaction completely.

The formation of bromate is an important reaction at high pH, but in neatral or acidic solutions it hardly occurs and it would require contact of the bromine with water for weeks before a noticeable loss occurs.

violet sin - 21-8-2013 at 00:54

So does the dry reflux of crude BrCl + Br over dry NaBr work? I didn't see a response to that above. Normally I would look it up but I'm stuck with an iTouch for a couple weeks and the tiny letters kill my eyes. I was wondering if there was enough kinetic energy to push that at reflux? Sorry for the laziness there. I'm not doing the process or any thing just curious. Thanks
-violet sin-

woelen - 21-8-2013 at 06:37

I am not sure whether this will work. NaBr is a solid and I'm quite sure that it will not dissolve in Br2 (or just a tiny little bit). Normally, you make a very concentrated solution of NaBr in water and you shake your impure Br2 with this solution. I would take 1 ml of solution of highly concentrated solution of NaBr or KBr for each 10 ml of bromine and shake this solution with the bromine vigorously to get it well mixed. Next step is to allow the layers to separate and separate the bromine from the aqueous layer. Losses will be low when only a small amount of solution of NaBr is used (keep in mind that Br2 dissolves quite well in a solution containing bromide ions). The bromine now contains some water. This can be removed by shaking it with conc. H2SO4 and separate the layers again. Even better purity is reached by distilling the bromine from the H2SO4/Br2 layers.