When KBr is treated with concentrated H_2SO_4 | vedclass

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(C) When $KBr$ reacts with concentrated $H_2SO_4$,$HBr$ is initially formed: $KBr + H_2SO_4 \rightarrow KHSO_4 + HBr$ $HBr$ is a strong reducing agent

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Bromine

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(C) When $KBr$ reacts with concentrated $H_2SO_4$,$HBr$ is initially formed: $KBr + H_2SO_4 \rightarrow KHSO_4 + HBr$ $HBr$ is a strong reducing agent and reduces concentrated $H_2SO_4$ to $SO_2$,while $HBr$ itself gets oxidized to reddish-brown bromine $(Br_2)$ gas: $2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$ Thus,the reddish-brown gas evolved is $Br_2$.

When $KBr$ is treated with concentrated $H_2SO_4$,a reddish-brown gas is evolved. The gas is:

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