In the reaction,BrO_{3(aq)}^{-} + 5Br_{(aq)}^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} 	o 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:Rate $= -\frac{d[BrO_3^

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} 	o 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:Rate $= -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = -\frac{1}{6} \frac{d[H^+]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$.We need to relate the rate of appearance of $Br_2$ to the rate of disappearance of $Br^-$.From the expression: $\frac{1}{3} \frac{d[Br_2]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt}$.Multiplying both sides by $3$,we get:$\frac{d[Br_2]}{dt} = -\frac{3}{5} \frac{d[Br^-]}{dt}$.

In the reaction,$BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} \to 3Br_{2(l)} + 3H_2O_{(l)}$. The rate of appearance of bromine $(Br_2)$ is related to the rate of disappearance of bromide ions as:

Similar Questions

Consider the graph given in the following diagram. Which of the following options does not show the instantaneous rate of reaction at the $40^{th}$ second?

If the first order reaction involves gaseous reactants and gaseous products,the units of its rate are:

If the rate of disappearance of $N_2O_5$ in the following reaction is $1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,the rate of production of $NO_2$ in $mol \ L^{-1} \ s^{-1}$ is:
$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module