For the reaction $2A + B \to A_2B$,the rate of reaction with respect to $A$ is $3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$. Calculate the rate of consumption of $B$ and the rate of formation of $A_2B$.
For the reaction $2A + B \to A_2B$,the rate expression is given by:
$Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt}$
Given that the rate of reaction with respect to $A$ is $-\frac{d[A]}{dt} = 3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$1$. Rate of consumption of $B$ $(-\frac{d[B]}{dt})$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = \frac{1}{2} \times 3.9 \times 10^{-9} = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$2$. Rate of formation of $A_2B$ $(\frac{d[A_2B]}{dt})$:
$\frac{d[A_2B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt}$
Given that the rate of reaction with respect to $A$ is $-\frac{d[A]}{dt} = 3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$1$. Rate of consumption of $B$ $(-\frac{d[B]}{dt})$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = \frac{1}{2} \times 3.9 \times 10^{-9} = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$2$. Rate of formation of $A_2B$ $(\frac{d[A_2B]}{dt})$:
$\frac{d[A_2B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
Explore More
Similar Questions
Nitric oxide reacts with $H_2$ according to the reaction: $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_2 O_{(g)}$. Identify the correct relation for the rate of disappearance of reactants and the rate of appearance of products.
The concentration of a reactant changes from $0.1 \, mol \, L^{-1}$ to $0.095 \, mol \, L^{-1}$ in $10 \, s$. Calculate the rate of reaction.
Medium
View SolutionFor the reaction $4KClO_3 \to 3KClO_4 + KCl$,if $-d[KClO_3]/dt = K_1[KClO_3]^4$,$d[KClO_4]/dt = K_2[KClO_3]^4$,and $d[KCl]/dt = K_3[KClO_3]^4$,then:
Difficult
View SolutionIn the following reaction; $x A \longrightarrow y B$
$\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$
$'A'$ and $'B'$ respectively can be
$\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$
$'A'$ and $'B'$ respectively can be
DifficultJEE MAIN 2019
View SolutionFill in the blanks:
$(a)$ The rate of reaction ....... as the concentration of reactant decreases.
$(b)$ The rate of reaction ....... as the concentration of product increases.
$(a)$ The rate of reaction ....... as the concentration of reactant decreases.
$(b)$ The rate of reaction ....... as the concentration of product increases.
Medium
View SolutionVedclass 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