Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

\mathrm{A} \rightarrow \mathrm{B}
The rate constants of the above reaction at 200 \mathrm{~K}$ and $300 \mathrm{~K} are 0.03 \mathrm{~min}^{-1} and 0.05 \mathrm{~min}^{-1} respectively. The activation energy for the reaction is J (Nearest integer) (Given : \ln 10=2.3 \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\log 5=0.70
\log 3=0.48
\log 2=0.30)

Option: 1

2520


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\operatorname{In}\left(\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}\right)=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]

\log \left(\frac{0.05}{0.03}\right)=\frac{E a}{2.3 \times 8.3}\left[\frac{1}{200}-\frac{1}{300}\right]
[0.70-0.48]=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{300-200}{300 \times 200}\right]
0.22=\frac{E a}{2.3 \times 8.3}\left[\frac{1}{600}\right]

\mathrm{Ea}=0.22 \times 2.3 \times 8.3 \times 600
       =2519.88 \mathrm{~J}
       =2519.88 \mathrm{~J}
       = 2520   

Posted by

Nehul

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

JEE Main 2026: NTA directs non-Aadhaar applicants to submit photo verification certificate by January 7
anam-khan

Jharkhand CM inaugurates free coaching institute for tribal medical, engineering students
anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

Latest Question