Br_2 + F_2 (Excess) xrightarrow{Delta} BrF_5. | vedclass

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(B) The reaction between bromine and fluorine depends on the molar ratio of the reactants.When $Br_2$ reacts with an excess of $F_2$,the product forme

Vedclass · Accepted Answer

$BrF_3$

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(B) The reaction between bromine and fluorine depends on the molar ratio of the reactants.When $Br_2$ reacts with an excess of $F_2$,the product formed is $BrF_5$.However,when $F_2$ is diluted with an inert gas like $N_2$,the concentration of $F_2$ is effectively reduced.Under these conditions,the reaction favors the formation of the lower interhalogen compound,which is $BrF_3$ $(Br_2 + 3F_2 \rightarrow 2BrF_3)$.

$Br_2 + F_2$ (Excess) $\xrightarrow{\Delta} BrF_5$. When $F_2$ is diluted with $N_2$ gas,then the product of $Br$ would be:

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