Write the reaction equations: HBr → NaBr → Br2 → HBr

1) Formation of sodium bromide during the interaction of acid and metal:

2HBr + 2Na = 2NaBr + H2.

2) Release of bromine as a result of electrolysis:

2NaBr + 2H2O = 2NaOH + H2 + Br2.

Possibly by the displacement reaction by the higher halogens:

2NaBr + Cl2 = 2NaCl + Br2.

3) Obtaining hydrogen bromide by fusion of simple substances at 200-400 ° C:

Br2 + H2 = 2HBr.

Or by the bromine reduction reaction:

Br2 + KNO2 + H2O = 2HBr + KNO3.