Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

4.30   The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Answers (1)

best_answer

From the Arrhenius equation,

\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303R}(\frac{T_{2}-T_{1}}{T_{1}T_{2}})...................................(i)
it is given that k_{2}=4k_{1}
T1= 293 K

T2 = 313 K
Putting all these values in equation (i) we get,

\log 4 =\frac{E_{a}}{2.303 \times 8.314}(\frac{313-293}{313 \times 293})

Activation Energy = 52.86 KJ/mol  
This is the required activation energy

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question