[painkiller]

Hi !

I have trouble figuring out the products of this reaction
NaBr + H₂SO₄ =

I know that Br^- will reduse H₂SO₄ to SO₂ but what about Na⁺ ions ?
Will it form a Na₂SO₄ ?

Thanks in advance

[AWK]

This reaction takes place with over 70 % H2SO4
NaBr + H2SO4 = Na2SO4 + Br2 + H2O + SO2

Balance this reaction

[painkiller]

Thank you ! you are my savior 

[woelen]

The reaction, posted by AWK indeed takes place, but I would not say that Na2SO4 is formed, but NaHSO4 (but that only is a minor detail).

More important, there also is another reaction and the majority of all bromide will end up with this:

H2SO4 + NaBr --> NaHSO4 + HBr

The reaction, mentioned by AWK is a side-reaction. So, in practice, when you add concentrated H2SO4 to NaBr, then HBr will be formed, which is contaminated with SO2 and Br2.

With NaCl instead of NaBr, only HCl is formed, the free halogen is not formed in that reaction.

With NaI instead of NaBr, the majority of the material is oxidized to iodine, SO2 and even H2S being formed from the acid. With NaI, only a small part of the iodide is converted to HI when concentrated sulphuric acid is added.

[AWK]

Yoa are right (at room temperature) and not right (at elevated temperatures)
Usually the temperatute is enough to remove bromine from reaction mixture bt distillation (over 60 C), and then my reaction is correct.

[woelen]

Your reaction is always correct, but there also always is another reaction. Both reactions occur at the same time. I can imagine that the higher the temperature, the more your reaction will occur, but I do not know of any situation, where only your reaction occurs, nor do I know of a situation, where only my reaction occurs.

This is a bad thing, because it makes this kind of reactions totally useless for preparative purposes. HBr cannot be made in a satisfactory way with this reaction, nor bromine.

[AWK]

This is a first step of my redox reaction.
2NaBr + 3H2SO4 = 2NaHSO4 + Br2 + 2H2O + SO2

The next step is:
2NaBr + 2NaHSO4  + H2SO4 = 2Na2SO4 + Br2 + 2H2O + SO2

When bromine is removed from reaction mixture (over boiling point of bromine) this reaction comes to completion (sum of both reactions)
4NaBr + 4H2SO4 = 2Na2SO4 + 2Br2 + 4H2O + 2SO2
and finally
2NaBr + 2H2SO4 = Na2SO4 + Br2 + 2H2O + SO2