What is the rate of formation of $O_2$ for the reaction stated below?
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
- A$0.01 \ mol \ dm^{-3} \ s^{-1}$
- B$0.02 \ mol \ dm^{-3} \ s^{-1}$
- C$0.03 \ mol \ dm^{-3} \ s^{-1}$
- D$0.04 \ mol \ dm^{-3} \ s^{-1}$
Explore More
Similar Questions
How is $r_{inst}$ obtained from a graph? Write the method to obtain $r_{inst}$.
Easy
View SolutionFor the reaction,$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$,$NH_{3}$ is formed at a rate of $0.088 \ mol \ dm^{-3} \ s^{-1}$. Calculate the consumption rate of $N_{2(g)}$.
In the following reaction: $xA \to yB$,$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$,where the negative sign indicates the rate of disappearance of the reactant. Thus,$x : y$ is:
Find the rate of the following reaction: $2 \ N_2O_{5(g)} \rightarrow 4 \ NO_{2(g)} + O_{2(g)}$ if the concentration of $NO_2$ increases to $5.2 \times 10^{-3} \ M$ in $100 \ s$.
For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the rate of reaction is expressed as:
MediumAIEEE 2002
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