Question

Concentrated sulphuric acid cannot be used to prepare pure $ HBr $ from $ NaBr $ as it reacts slowly with $ NaBr $ .
A. True
B. False

Answer 1

Hint: It is true that concentrated sulphuric acid cannot be used to prepare pure $ HBr $ from $ NaBr $ but its given reason is not true. It does not react slowly with $ NaBr $ . The reason is the reducing power of hydrogen bromide. It reduces the acid and oxidizes itself. Due to which this statement is false.

Complete step by step solution
Hydrogen bromide is a heteronuclear diatomic molecular compound having chemical formula as $ HBr $ . It is colorless and forms hydrobromic acid when dissolved in water. Simply it consists of hydrogen and bromine atoms. It is a colorless gas in pure form. And it is also a very strong reducing agent. It is highly corrosive in nature and it is also very irritating to inhale.
Hydrogen bromide can be prepared by a variety of methods. It can be prepared by the distillation of a solution of sodium bromide or potassium bromide with phosphoric acid. As shown in the reaction:
 $ KBr + {H_2}S{O_4} \to KHS{O_4} + HBr $
But concentrated sulphuric acid is not used here because hydrogen bromide is a strong reducing agent and it reduces sulphuric acid to sulphur dioxide and oxidises itself and forms a molecule of bromine.
Reaction will be:
 $ {H_2}S{O_4} + NaBr \to NaHS{O_4} + HBr $
 $ {H_2}S{O_4} + 2HBr \to S{O_2} + B{r_2} + 2{H_2}O $
Hence, option B is correct.

Note
When Hydrogen bromide get prepared then it can be easily contaminated with dibromine that is $ B{r_2} $ , that can be removed by passing the gas through a solution of phenol at room temperature in tetrachloromethane or other suitable solvent (which produce 2,4,6-tribromophenol and generating more HBr in the process) or through copper turnings or also through copper gauze at high temperature. The anhydrous solution and aqueous solutions of hydrogen bromide are common reagents in the preparation of bromide compounds.