Question

Concentrated ${{H}_{2}}S{{O}_{4}}$ cannot be used to prepare $HBr$ from $NaBr$ because it _______________.
A. Reacts slowly with $NaBr$
B. Oxidises $HBr$
C. Reduces $HBr$
D. Disproportionates $HBr$

Answer 1

Hint: In the given question, we have to answer why concentrated sulphuric acid cannot be used to prepare pure HBr from $NaBr$. For this, we must know about the properties of ${{H}_{2}}S{{O}_{4}}$ and $HBr$. Let us analyse some of the points to give the correct answer.

Complete Step by Step Solution:
Hydrogen bromide is a heteronuclear diatomic molecular compound which has a chemical formula as $HBr$ . It is colourless and it forms hydrobromic acid when dissolved in water. It is made of hydrogen and bromine atoms. It is a colourless gas in pure form and acts as a very strong reducing agent. It is highly corrosive in nature and it is also very irritating to breathe.

Hydrogen bromide can be prepared by many methods. It can be prepared by the distillation of a solution of sodium bromide or potassium bromide with phosphoric acid which is 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 itself 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:
$NaBr+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HBr$
$2HBr+{{H}_{2}}S{{O}_{4}}\to S{{O}_{2}}+B{{r}_{2}}+2{{H}_{2}}O$
Thus, Option (B) is correct.

Note: Here to answer these types of questions, we must have the proper knowledge of the properties of the chemicals which are given in the question. After that we are able to answer the question correctly.