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Concentrated \mathrm{H}_{2} \mathrm{SO}_{4} cannot be used to prepare \mathrm{HBr} from \mathrm{NaBr},Because it:

Option: 1

Reduces \mathrm{HBr}


Option: 2

oxidises \mathrm{HBr}


Option: 3

disproportionate \mathrm{HBr}


Option: 4

reacts slowly with \mathrm{NaBr}


Answers (1)

best_answer

\mathrm{HBr} is Reducing Agent and \mathrm{H}_{2} \mathrm{SO}_{4} is Oxidising Agent and thus, action of \mathrm{NaBr} with \mathrm{H}_{2} \mathrm{SO}_{4} gives bromine in place of \mathrm{HBr}.

\mathrm{NaBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{NaHSO}_{4}+\mathrm{HBr}
\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HBr} \rightarrow \mathrm{SO}_{2}+\mathrm{Br}_{2}+2 \mathrm{H}_{2} \mathrm{O}

Posted by

shivangi.bhatnagar

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