(D) For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t}$ Given $[X]_0 = 1.0 \ M$,$[X]_t = 0.25

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

$3.45 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$

(D) For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t}$ Given $[X]_0 = 1.0 \ M$,$[X]_t = 0.25 \ M$,and $t = 40 \ min$: $k = \frac{2.303}{40} \log \frac{1.0}{0.25} = \frac{2.303}{40} \log 4$ $k = \frac{2.303 \times 0.60}{40} = 0.034545 \ min^{-1}$ Now,the rate of reaction when $[X] = 0.1 \ M$ is: $Rate = k[X] = 0.034545 \times 0.1$ $Rate = 0.0034545 \ mol \ L^{-1} \ min^{-1} = 3.45 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$

Class 12 Chemistry Chemical Kinetics First Order reaction

Difficult

The reaction $X \rightarrow$ products is a first order reaction. In $40 \ min$,the concentration of $X$ changes from $1.0 \ M$ to $0.25 \ M$. What is the rate of reaction when $[X] = 0.1 \ M$? $(\log 4 = 0.60)$

AP EAMCET 2008 All AP EAMCET

AP EAMCET 2022 All AP EAMCET

A
$1.73 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$
B
$3.47 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}$
C
$1.73 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}$
D
$3.45 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

First Order reaction

Previous Year

AP EAMCET 2008 Paper All AP EAMCET years →

Previous Year

AP EAMCET 2022 Paper All AP EAMCET years →

After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

MediumAIPMT 2000

View Solution

Time required for $99.9 \%$ completion of a first order reaction is . . . . . . . times the time required for completion of $90 \%$ reaction. (nearest integer).

MediumJEE MAIN 2024

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

At $300 \ K$,a gaseous reaction $A(g) \longrightarrow B(g) + C(g)$ follows first-order kinetics. Starting with pure $A$,the total pressure at the end of $20 \ min$ is $100 \ mm \ of \ Hg$. The total pressure after the completion of the reaction is $180 \ mm \ of \ Hg$. The partial pressure of $A$ at $20 \ min$ (in $mm \ of \ Hg$) is:

MediumKCET 2012

View Solution

For the reaction $A \rightarrow B$,the following graph was obtained. The time required (in seconds) for the concentration of $A$ to reduce to $2.5 \ g \ L^{-1}$ (if the initial concentration of $A$ was $50 \ g \ L^{-1}$) is $........$ (Nearest integer). Given: $\log 2 = 0.3010$.

DifficultJEE MAIN 2025

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 reaction $X \rightarrow$ products is a first order reaction. In $40 \ min$,the concentration of $X$ changes from $1.0 \ M$ to $0.25 \ M$. What is the rate of reaction when $[X] = 0.1 \ M$? $(\log 4 = 0.60)$

Similar Questions

After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

Time required for $99.9 \%$ completion of a first order reaction is . . . . . . . times the time required for completion of $90 \%$ reaction. (nearest integer).

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 the reaction $A \rightarrow B$,the following graph was obtained. The time required (in seconds) for the concentration of $A$ to reduce to $2.5 \ g \ L^{-1}$ (if the initial concentration of $A$ was $50 \ g \ L^{-1}$) is $........$ (Nearest integer). Given: $\log 2 = 0.3010$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module