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$H_{2} {SO}_{4}$ cannot be used to prepare $HBr$ from $NaBr$ as it

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Oxidises $HBr$

Reacts slowly with $NaBr$

Reduces $HBr$

Disproportionate $HBr$

answer is B.

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Detailed Solution

$HBr$ is not prepared by heating $NaBr$ with conc. $H_{2} {SO}_{4}$ because $HBr$ is a strong reducing agent and reduce $H_{2} {SO}_{4}$ to ${SO}_{2}$ and get itself oxidised to bromine.

$\begin{array}{l} NaBr + H_{2} {SO}_{4} \to {NaHSO}_{4} + HBr \\ H_{2} {SO}_{4} + 2 HBr \to {SO}_{2} + {Br}_{2} + 2 H_{2} O \end{array}$

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