Good questions! Now we're getting somewhere. Okay, so what are we looking at as far as equilibria? Well, under aqueous conditions, with a mixture of Br2 and HBr, we have the following:
H-Br (hydrogen bromide) + H-O-H (H2O) <-> H3O+ (hydronium anion) + Br- (bromide cation)
Br-Br (elemental bromine) + H-O-H (H2O) <-> H-Br (hydrogen bromide) + HOBr (hypobromic acid)
and, we have the following reaction:
2 HOBr + (UV light) -> 2 HBr + O2
Now, when you add bromine to water, you get an equilibrium, where Bromine dissociates and reacts with one equivalent of water, forming one equivalent of hydrobromic acid, and one equivalent of hypobromic acid. After awhile, the HBr and HOBr react with each other, and the product is H2O and Br2. This goes on and on, back and forth, countless times in the solution, and its a relatively stable pattern.
Now if you shine some strong UV light source on the solution (like sunlight), something else happens. The light causes an irreversible reaction to happen,so that O2 gas and HBr evolve. Since the solubility of O2 in solution is only about 9 mg/L, it bubbles off quite quickly. However, like it was said before, a saturated HBr solution is 66% HBr, meaning you'd have to produce a hell of a lot of HBr before the solution wouldn't hold any more, and then you'd still have to separate HBr from the Br2 fumes. Also, you'd need a lot of light, or a lot of time, or both. In theory at least, it should work, but it sure sounds messy.
Hey, here's an idea:
NaBH4 + 4Br2 + 3H2O -> 7HBr + NaBr + B(OH)3
Now, to keep it relatively anhydrous (not completely anhydrous per se, since 3 equilvalents of water are used), we need a relatively innert solvent that will dissolve the starting materials well, but not the final products (well, at least not dissolve the HBr well, anyway.) Glyme or diglyme or even THF sound like fine candidates. If everything is kept relatively cold, perhaps by chilling the NaBH4/glyme solution using ice and rock salt, then one could add elemental bromine with a pressure-equalized addition funnel. Perhaps by adding the bromine and water in series, the borohydride wouldn't have to compete for solvation against boric acid. Hmmm...
Step 1:
NaBH4 + 4Br2 -> 4HBr + NaBr + BBr3
Step 2:
BBr3 + 3H2O -> 3HBr + B(OH)3
It looks like this might work, but I have a funny feeling I'm overlooking something somewhere. To the other killer bees: does this look reasonable? Am I somehow forgetting some crucial side-reaction? I mean, this isn't all that great of an alternative, but would this work?
To Agent_Smith: I just wanted to appologise again too, in case I sounded rude in my last response. I have to admit, sometimes I forget my manners. I'm sure you know that we're all on the same side of this terrible war. I took the time to give a thoughtful answer, which I genuinely hope you enjoy and find useful. I'd hate for that effort to go unappreciated, simply because of some of my bad choices in phrasology.