$A$ complex reaction takes place in $2$ steps:
$(I)$ $NO_{2(g)} + F_{2(g)} \rightarrow NO_2F_{(g)} + F_{(g)}$ (slow)
$(II)$ $F_{(g)} + NO_{2(g)} \rightarrow NO_2F_{(g)}$ (fast)
Identify the reaction intermediate.
$(I)$ $NO_{2(g)} + F_{2(g)} \rightarrow NO_2F_{(g)} + F_{(g)}$ (slow)
$(II)$ $F_{(g)} + NO_{2(g)} \rightarrow NO_2F_{(g)}$ (fast)
Identify the reaction intermediate.
- A$F_{(g)}$
- B$NO_{2(g)}$
- C$F_{2(g)}$
- D$NO_2F_{(g)}$
Explore More
Similar Questions
For an $n^{th}$ order reaction,the rate law is given by $\frac{dx}{dt} = K[A]^n$. Which parameter can be determined from the logarithmic plot?
Medium
View SolutionThe rate law of the reaction $2N_2O_5 \to 4NO_2 + O_2$ is
Easy
View SolutionCalculate the order of the reaction with respect to $A$ and $B$ based on the following data:
$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$ $0.05$ $0.05$ $1.2 \times 10^{-3}$ $0.10$ $0.05$ $2.4 \times 10^{-3}$ $0.05$ $0.10$ $1.2 \times 10^{-3}$
| $[A] \ (mol/L)$ | $[B] \ (mol/L)$ | Rate $(mol/L \cdot s)$ |
| $0.05$ | $0.05$ | $1.2 \times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4 \times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2 \times 10^{-3}$ |
Easy
View SolutionFor a reaction between $A$ and $B$,the order with respect to $A$ is $2$ and the order with respect to $B$ is $3$. If the concentrations of both $A$ and $B$ are doubled,the rate will increase by a factor of:
MediumNEET 2013
View Solution$A + B \rightarrow \text{product}$. The rate of this reaction is given by $v = K[A]^2[B]^0$. What is the change in the rate of reaction when the concentration of $A$ is doubled and the concentration of $B$ is doubled?
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