The decomposition of $N_2O_5$ is a first order reaction represented by $N_2O_5 \to N_2O_4 + \frac{1}{2} O_2$. After $15 \ min$,the volume of $O_2$ produced is $9 \ mL$ and at the end of the reaction,it is $35 \ mL$. The rate constant is equal to:

  • A
    $\frac{1}{15} \ln \frac{35}{44}$
  • B
    $\frac{1}{15} \ln \frac{44}{26}$
  • C
    $\frac{1}{15} \ln \frac{44}{35}$
  • D
    $\frac{1}{15} \ln \frac{35}{26}$

Explore More

Similar Questions

The time taken for $10 \%$ completion of a first order reaction is $20$ minutes. The time required for the completion of $19 \%$ of the same reaction in minutes is

If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........
(Take: $\log 2=0.30 ; \log 2.5=0.40)$

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.

For a first-order reaction $A \rightarrow \text{Product}$,the rate of reaction is $1 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}$ when $[A] = 0.2 \, \text{M}$. What is the half-life $(t_{1/2})$ of the reaction?

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