The differential rate law for the reaction $H_2 + I_2 \to 2HI$ is
- A$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = +\frac{1}{2}\frac{d[HI]}{dt}$
- B$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = +\frac{1}{2}\frac{d[HI]}{dt}$
- C$\frac{1}{2}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[I_2]}{dt} = -\frac{d[HI]}{dt}$
- D$-2\frac{d[H_2]}{dt} = -2\frac{d[I_2]}{dt} = +\frac{d[HI]}{dt}$
Explore More
Similar Questions
For the reaction $2A + 3B \rightarrow 4C$,the rate of reaction is expressed as:
Easy
View SolutionIn a reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of appearance of $NH_3$ is $2.5 \times 10^{-4} \ mol \ L^{-1} s^{-1}$. The rate of reaction and rate of disappearance of $H_2$ will be (in $mol \ L^{-1} s^{-1}$):
Difficult
View SolutionIn a reaction $R \to P$,the concentration $[M]$ obtained at different times $(t)$ are shown in the following table. Calculate the average rate $r_{av}$ of the reaction.
Time $t$ $(s)$ $0$ $5$ $10$ $20$ $30$ Concentration $(mol \ L^{-1})$ $160 \times 10^{-3}$ $80 \times 10^{-3}$ $40 \times 10^{-3}$ $10 \times 10^{-3}$ $3.5 \times 10^{-3}$
| Time $t$ $(s)$ | $0$ | $5$ | $10$ | $20$ | $30$ |
| Concentration $(mol \ L^{-1})$ | $160 \times 10^{-3}$ | $80 \times 10^{-3}$ | $40 \times 10^{-3}$ | $10 \times 10^{-3}$ | $3.5 \times 10^{-3}$ |
Medium
View SolutionFor the reaction $CH_3COOCH_3 + H_2O \xrightarrow{H^{+}} CH_3COOH + CH_3OH$,the progress of the reaction is followed by:
Easy
View SolutionIf the first order reaction involves gaseous reactants and gaseous products,the units of its rate are:
Easy
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