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Chemistry Question on Chemical Kinetics

In the reaction BrO3−(aq)+5Br−(aq)+6H+→3Br2(l)+3H2O(l)BrO _{3}^{-}( aq )+5 Br ^{-}( aq )+6 H ^{+} \rightarrow 3 Br _{2}(l)+3 H _{2} O (l) The rate of appearance of bromine (Br2)\left( Br _{2}\right) is related to rate of disappearance of bromide ions as following

A

d[Br2]dt=−35d[Br−]dt\frac{d [Br_2]}{dt} = - \frac{3}{5} \frac{d [Br^-]}{dt}

B

d[Br2]dt=−53d[Br−]dt\frac{d [Br_2]}{dt} = - \frac{5}{3} \frac{d [Br^-]}{dt}

C

d[Br2]dt=53d[Br−]dt\frac{d [Br_2]}{dt} = \frac{5}{3} \frac{d [Br^-]}{dt}

D

d[Br2]dt=35d[Br−]dt\frac{d [Br_2]}{dt} =\frac{3}{5} \frac{d [Br^-]}{dt}

Answer

d[Br2]dt=−35d[Br−]dt\frac{d [Br_2]}{dt} = - \frac{3}{5} \frac{d [Br^-]}{dt}

Explanation

Solution

The correct option is(A): d[Br2]dt=−35d[Br−]dt\frac{d [Br_2]}{dt} = - \frac{3}{5} \frac{d [Br^-]}{dt}.

−14d(Br−)dt=−13d(Br2)dt-\frac{1}{4} \frac{ d \left( Br ^{-}\right)}{ dt }=-\frac{1}{3} \frac{ d \left( Br _{2}\right)}{ dt }
d(Br2)dt=−35d(Br−)dt\frac{ d \left( Br _{2}\right)}{ dt }=-\frac{3}{5} \frac{ d \left( Br ^{-}\right)}{ dt }