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In the reaction BrO_{3}^{-}_{(aq)}+5Br^{-}_{(aq)}+6H^{+}\rightarrow 3Br_{2}_{(l)}+3H_{2}O(l)

The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as following:

 

  • Option 1)

    \frac{d\left [ Br_{2} \right ]}{dt}=-\frac{5}{3}\frac{d\left [ Br^{-} \right ]}{dt}

  • Option 2)

    \frac{d\left [ Br_{2} \right ]}{dt}=\frac{5}{3}\frac{d\left [ Br^{-} \right ]}{dt}

  • Option 3)

    \frac{d\left [ Br_{2} \right ]}{dt}=\frac{3}{5}\frac{d\left [ Br^{-} \right ]}{dt}

  • Option 4)

    \frac{d\left [ Br_{2} \right ]}{dt}=-\frac{3}{5}\frac{d\left [ Br^{-} \right ]}{dt}

 

Answers (2)

 

Rates in presence of stoichiometry of reactants/products -

When stoichiometry coefficients of reactants/ products are not equal to one, the rate of disappearance of  & the rate of appearance of products is divided by their respective stoichiometric coefficients

- wherein

e.g.\:2HI(g)\rightarrow H_{2}(g)+I_{2}(g)

r=\frac{-1}{2}.\frac{d}{dt}[HI]

=\frac{+d}{dt}[H2]=\frac{+ d}{dt}[I_{2}]

 

 BrO_{3}^{-}_{\left ( aq \right )}+5Br^{-}_{\left ( aq \right )}+6H^{+}\rightarrow 3Br_{2}_{\left ( l \right )}+3H_{2}O_{\left ( l \right )}

R=\frac{-d[BrO_{3}^{-}]}{dt}=-\frac{1}{5}\frac{d[Br^{-}]}{dt}=-\frac{1}{6}\frac{d[H^{+}]}{dt}

\Rightarrow \frac{1}{3}\frac{d[Br_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}O]}{dt}

\Rightarrow -\frac{1}{5}\frac{d[Br^{-}]}{dt}=\frac{1}{3}\frac{d[Br_2]}{dt}

\Rightarrow \frac{d[Br_{2}]}{dt}=-\frac{3}{5}\frac{d[Br^{-}]}{dt}

 


Option 1)

\frac{d\left [ Br_{2} \right ]}{dt}=-\frac{5}{3}\frac{d\left [ Br^{-} \right ]}{dt}

This is incorrect option

Option 2)

\frac{d\left [ Br_{2} \right ]}{dt}=\frac{5}{3}\frac{d\left [ Br^{-} \right ]}{dt}

This is incorrect option

Option 3)

\frac{d\left [ Br_{2} \right ]}{dt}=\frac{3}{5}\frac{d\left [ Br^{-} \right ]}{dt}

This is incorrect option

Option 4)

\frac{d\left [ Br_{2} \right ]}{dt}=-\frac{3}{5}\frac{d\left [ Br^{-} \right ]}{dt}

This is correct option

Posted by

Vakul

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As the rate of disappearance of reactants is equal to the rate of formation of the products,there should be a  minus sign with the coefficients of the required substances in the denominator.As in this case,\frac{d\left [ Br_{2} \right ]}{dt}=-\frac{3}{5}\frac{d\left [ Br^{-} \right ]}{dt}  is the right answer as 3 goes th the lhs in the denominator.

Posted by

BHARATH CHARAN

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