For the reaction ${N_2}{O_{5(g)}} \to 2N{O_{2(g)}} + \frac{1}{2}{O_{2(g)}}$,the rate constant is $2.3 \times 10^{-2} \ s^{-1}$. Which of the following equations represents the variation of $[{N_2}{O_5}]$ with time?

  • A
    $[{N_2}{O_5}]_t = [{N_2}{O_5}]_0 \ e^{-Kt}$
  • B
    $\ln \frac{[{N_2}{O_5}]_0}{[{N_2}{O_5}]_t} = Kt$
  • C
    $\log_{10} [{N_2}{O_5}]_t = \log_{10} [{N_2}{O_5}]_0 - \frac{Kt}{2.303}$
  • D
    $[{N_2}{O_5}]_t = [{N_2}{O_5}]_0 + Kt$

Explore More

Similar Questions

For the $1^{st}$ order reaction,the half-life is $5 \ minutes$ when $[A] = 0.1 \ M$. If the concentration of $[A]$ becomes twice,then the half-life becomes:

The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $1/20^{th}$ of its initial value (in $\text{ s}$)?

The integrated rate equation for a first-order reaction,$A \rightarrow \text{product}$,is

For a first-order reaction,the half-life period is $69.3 \ s$. If the concentration of the reactant is $0.10 \ mol \ L^{-1}$,what will be the rate of the reaction?

The rate of a first-order reaction is $0.04 \ mol \ L^{-1} \ s^{-1}$ at $10 \ s$ and $0.03 \ mol \ L^{-1} \ s^{-1}$ at $20 \ s$ after the initiation of the reaction. The half-life period of the reaction is ......... $s$. (in $.1$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo