For the reaction $5Br_{(aq)}^{-} + BrO_{3_{(aq)}}^{-} + 6H_{(aq)}^{+} \rightarrow 3Br_{2_{(aq)}} + 3H_{2}O_{(l)}$,if $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,calculate the rate of formation of $Br_{2}$,i.e.,$\frac{\Delta[Br_{2}]}{\Delta t}$.
$(2.52 \times 10^{-3} \ MOL \ L^{-1} \ S^{-1})$ From the stoichiometry of the balanced chemical equation,the rate of reaction is given by:
$\frac{-1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
Given that $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,we substitute this into the expression:
$\frac{1}{5} (4.2 \times 10^{-3}) = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
$\frac{\Delta[Br_{2}]}{\Delta t} = \frac{3}{5} \times 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{\Delta[Br_{2}]}{\Delta t} = 2.52 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{-1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
Given that $\frac{-\Delta[Br^{-}]}{\Delta t} = 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,we substitute this into the expression:
$\frac{1}{5} (4.2 \times 10^{-3}) = \frac{1}{3} \frac{\Delta[Br_{2}]}{\Delta t}$
$\frac{\Delta[Br_{2}]}{\Delta t} = \frac{3}{5} \times 4.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
$\frac{\Delta[Br_{2}]}{\Delta t} = 2.52 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
Explore More
Similar Questions
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:
$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$
$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$
At $298 \ K$ the value of $-\frac{\Delta[Br^{-}]}{\Delta t}$ for the reaction$5Br^-{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$ is $x \ mol \ L^{-1} \ min^{-1}$. What is the rate (in $mol \ L^{-1} \ min^{-1}$) of this reaction?
For the reaction $A + 2B \to C + D$,if $-\frac{d[A]}{dt} = 5 \times 10^{-4} \ mol \ L^{-1} s^{-1}$,then what is the value of $-\frac{d[B]}{dt}$?
Medium
View SolutionFor the decomposition of $A$ as shown below,calculate the rate of appearance of $B$ if the initial concentration of $A$ is $2 \ M$.
$A_{(g)} \xrightarrow{K_1 = 2 \times 10^{-3} \ s^{-1}} 2B_{(g)}$
$A_{(g)} \xrightarrow{K_2 = 1 \times 10^{-3} \ s^{-1}} C_{(g)}$
$A_{(g)} \xrightarrow{K_1 = 2 \times 10^{-3} \ s^{-1}} 2B_{(g)}$
$A_{(g)} \xrightarrow{K_2 = 1 \times 10^{-3} \ s^{-1}} C_{(g)}$
Medium
View SolutionFor a chemical reaction $A \rightarrow B$,it was found that the concentration of $B$ increases by $0.2 \, mol \, L^{-1}$ in $30 \, min$. The average rate of the reaction is $...... \times 10^{-1} \, mol \, L^{-1} \, h^{-1}$. (Nearest integer)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo