(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0

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

$0.0693 \times 3 \ M \ min^{-1}$

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0693 \ min^{-1}$.After $20 \ min$ (which is $2 \times t_{1/2}$),the concentration $[A]$ will be reduced to $\frac{1}{4}$ of the initial concentration.$[A] = \frac{12}{4} = 3 \ M$.The rate of reaction is given by $Rate = k[A]$.$Rate = 0.0693 \times 3 \ M \ min^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Difficult

Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

KCET 2015 All KCET

A
$0.0693 \ M \ min^{-1}$
B
$0.693 \times 3 \ M \ min^{-1}$
C
$0.0693 \times 3 \ M \ min^{-1}$
D
$0.0693 \times 4 \ M \ min^{-1}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

First Order reaction

Previous Year

KCET 2015 Paper All KCET years →

If the concentration is expressed in $mol \ L^{-1}$,the unit of the rate constant for a first-order reaction is........

Easy

View Solution

Half-life of a first order reaction is $20 \ s$ and initial concentration of reactant is $0.2 \ M$. The concentration of reactant left after $80 \ s$ is (in $M$)

MediumKCET 2025

View Solution

In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

MediumAP EAMCET 2025

View Solution

The following results were obtained during kinetic studies of the reaction $2A + B \to$ products:

Experiment $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ min^{-1})$

$I$ $0.10$ $0.20$ $6.93 \times 10^{-3}$

$II$ $0.10$ $0.25$ $6.93 \times 10^{-3}$

$III$ $0.20$ $0.30$ $1.386 \times 10^{-2}$

The time (in minutes) required to consume half of $A$ is:

Experiment	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times 10^{-3}$
$II$	$0.10$	$0.25$	$6.93 \times 10^{-3}$
$III$	$0.20$	$0.30$	$1.386 \times 10^{-2}$

DifficultJEE MAIN 2019

View Solution

The time of completion of $90\%$ of a first order reaction is approximately

Medium

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

Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

Similar Questions

If the concentration is expressed in $mol \ L^{-1}$,the unit of the rate constant for a first-order reaction is........

Half-life of a first order reaction is $20 \ s$ and initial concentration of reactant is $0.2 \ M$. The concentration of reactant left after $80 \ s$ is (in $M$)

In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

The time of completion of $90\%$ of a first order reaction is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module