(C) For a first-order reaction,the rate $R$ at time $t$ is given by $R = k[A]_t$,where $[A]_t = [A]_0 e^{-kt}$.Thus,$R_t = k[A]_0 e^{-kt} = R_0 e^{-kt

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) For a first-order reaction,the rate $R$ at time $t$ is given by $R = k[A]_t$,where $[A]_t = [A]_0 e^{-kt}$.Thus,$R_t = k[A]_0 e^{-kt} = R_0 e^{-kt}$.Given $R_{10} = 0.04 \ mol \ L^{-1} \ s^{-1}$ and $R_{20} = 0.03 \ mol \ L^{-1} \ s^{-1}$.Taking the ratio: $\frac{R_{10}}{R_{20}} = \frac{e^{-k(10)}}{e^{-k(20)}} = e^{10k} = \frac{0.04}{0.03} = 1.333$.Taking natural log on both sides: $10k = \ln(1.333) = 0.2877$.$k = 0.02877 \ s^{-1}$.The half-life period is $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02877} \approx 24.1 \ s$.

Class 12 Chemistry Chemical Kinetics First Order 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 $.1$)

NEET 2016 All NEET

A
$44$
B
$54$
C
$24$
D
$34$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

First Order reaction

Previous Year

NEET 2016 Paper All NEET years →

For the reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,which is a first-order reaction with respect to $N_2O_5$,which of the following plots gives a straight line?

Medium

View Solution

The half-life of a first-order reaction is $10 \ minutes$. If the initial concentration is $0.08 \ mol/L$ and at some instant the concentration is $0.01 \ mol/L$,then the time elapsed is ...... minutes.

Easy

View Solution

An organic compound undergoes first order decomposition. If the time taken for the $60 \%$ decomposition is $540 \ s$,then the time required for $90 \%$ decomposition will be $.......... \ s$. (Nearest integer). Given : $\ln 10 = 2.3 ; \log 2 = 0.3$

DifficultJEE MAIN 2023

View Solution

For a first order reaction,show that the time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of the reaction.

Medium

View Solution

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

MediumJEE MAIN 2022

View Solution

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

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 $.1$)

Similar Questions

For the reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,which is a first-order reaction with respect to $N_2O_5$,which of the following plots gives a straight line?

The half-life of a first-order reaction is $10 \ minutes$. If the initial concentration is $0.08 \ mol/L$ and at some instant the concentration is $0.01 \ mol/L$,then the time elapsed is ...... minutes.

An organic compound undergoes first order decomposition. If the time taken for the $60 \%$ decomposition is $540 \ s$,then the time required for $90 \%$ decomposition will be $.......... \ s$. (Nearest integer). Given : $\ln 10 = 2.3 ; \log 2 = 0.3$

For a first order reaction,show that the time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of the reaction.

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module