Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • For a first order reaction, the slope of the graph between  <img alt="\mathrm{\ln _e\left([A]_0 / [A]\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cln%20_e%

For a first order reaction, the slope of the graph between  \mathrm{\ln _e\left([A]_0 / [A]\right)} (on y-axis) and time (min) (on x-axis) is found out to be \mathrm{0.5 min{^{-1}}} . The time at which 25 % of the initial reactant is remaining is:

Here,
\mathrm{[A]_0 } is the initial concentration
\mathrm{[A]} is the concentration at time t

Option: 1

1.38 min


Option: 2

2.77 min


Option: 3

0.693 min


Option: 4

0.5 min


Answers (1)

best_answer

For a first order:
\mathrm{[A]=[A]_0 e^{-k t}}
\mathrm{ \ln _e\left(\frac{[A]_0}{[A]}\right)=k t }
Comparing with

y=mx

Here slope is the rate constant.
So,

\mathrm{ k=0.5 \ min ^ {-1} }

At
\mathrm{ \ln _e\left(\frac{[A]_0}{[A]_0}\right)=k t \\ }

\mathrm{ \ln _e 4=0.5 \times t \\ }
\mathrm{ t=\frac{2 \times 0.693}{0.5} \mathrm{~min} \\ }

\mathrm{ t=2.77 \mathrm{~min} }.
 

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30
anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question