When KBr is treated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) a reddish-brown gas is evolved. The…

When KBr is treated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) a reddish-brown gas is evolved. The evolved gas is
  1. Bromine
  2. Mixture of bromine and \(\mathrm{HBr}\)
  3. \(\mathrm{HBr}\)
  4. \(\mathrm{NO}_{2}\)

Solution

$\mathrm{KBr}$, on reaction with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, gives reddish-brown bromine gas.
$\mathrm{KBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{KHSO}_{4}+\mathrm{HBr}, 2 \mathrm{HBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Br}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{2}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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