(A) For a first order reaction,the rate law is given by $Rate = k[R]$.The half-life period $(t_{1/2})$ for a first order reaction is derived as $t_{1/

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

(A) For a first order reaction,the rate law is given by $Rate = k[R]$.The half-life period $(t_{1/2})$ for a first order reaction is derived as $t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$.Since $t_{1/2}$ is independent of the initial concentration $[R]_0$,it can be expressed as $t_{1/2} \propto [R]_0^0$,which means $t_{1/2}$ is proportional to the initial concentration raised to the power of $0$.Both statements $(i)$ and $(ii)$ are correct. Therefore,the correct option is $(A)$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Easy

Identify True $(T)$ and False $(F)$ statements of the following for a first order reaction $R \rightarrow P$.
$(i) \ t_{1/2} \propto [R]_0^0$
$(ii) \ t_{1/2} = \frac{0.693}{k}$

A
$(i) T, (ii) T$
B
$(i) T, (ii) F$
C
$(i) F, (ii) T$
D
$(i) F, (ii) F$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

First Order reaction

The half-life for the reaction $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$ is $24 \ hr$ at $30 \ ^\circ C$. Starting with $10 \ g$ of $N_2O_5$,how many grams of $N_2O_5$ will remain after a period of $96 \ hr$?

Difficult

View Solution

Half life of a first order reaction is $20 \ min$. The time taken to reduce the initial concentration of reactant to $(1/10)^{th}$ is $\qquad$ (in $min$)

MediumMHT CET 2025

View Solution

If the half-life $(t_{1/2})$ of a reaction is $69.3 \ s$ and the rate constant $(k)$ is $10^{-2} \ s^{-1}$,then the order of the reaction is:

Easy

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

For the reaction $2A + B \to \text{product}$,the rate law is given by $\frac{-d[A]}{dt} = k[A]$. At a time $t = \frac{t_{1/2}}{\ln 2}$,what is the concentration of the reactant $A$?

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

Identify True $(T)$ and False $(F)$ statements of the following for a first order reaction $R \rightarrow P$.$(i) \ t_{1/2} \propto [R]_0^0$$(ii) \ t_{1/2} = \frac{0.693}{k}$

Similar Questions

The half-life for the reaction $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$ is $24 \ hr$ at $30 \ ^\circ C$. Starting with $10 \ g$ of $N_2O_5$,how many grams of $N_2O_5$ will remain after a period of $96 \ hr$?

Half life of a first order reaction is $20 \ min$. The time taken to reduce the initial concentration of reactant to $(1/10)^{th}$ is $\qquad$ (in $min$)

If the half-life $(t_{1/2})$ of a reaction is $69.3 \ s$ and the rate constant $(k)$ is $10^{-2} \ s^{-1}$,then the order of the reaction is:

For the reaction $2A + B \to \text{product}$,the rate law is given by $\frac{-d[A]}{dt} = k[A]$. At a time $t = \frac{t_{1/2}}{\ln 2}$,what is the concentration of the reactant $A$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify True $(T)$ and False $(F)$ statements of the following for a first order reaction $R \rightarrow P$.
$(i) \ t_{1/2} \propto [R]_0^0$
$(ii) \ t_{1/2} = \frac{0.693}{k}$