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Reaction with Metals

H_2 SO_4 cann...

Question

$H_{2} S O_{4}$ cannot be used to prepare HBr from NaBr as it:

A

Reacts slowly with NaBr

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B

Oxidises HBr

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C

Reduces HBr

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D

Disproportionate HBr

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Solution

The correct option is B Oxidises HBr
HBr is RA and $H_{2} S O_{4}$ is OA and thus , action of NaBr with $H_{2} S O_{4}$ gives bromine in place of HBr

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