Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • The rate constant of a first order reaction increases from  to <img alt="6\times 10^{-2}" src="https://entrancecorner.codecogs.com/gif.latex

The rate constant of a first order reaction increases from 2\times 10^{-2} to 6\times 10^{-2} when the temperature changes from 300 K to 320 K. Calculate the energy of activation. (Given : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

 

 

 

 
 
 
 
 

Answers (1)

Order = 1

\begin{aligned} &E_{a}=? \quad R=8.314 \mathrm{J} / \mathrm{K}\\ &T_{1}=300 K \quad k_{1}=2 \times 10^{-2}\\ &T_{2}=320 K \quad k_{2}=6 \times 10^{-2}\\ &\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]\\ &\log \frac{6 \times 10^{-2}}{2 \times 10^{-2}}=\frac{E_{a}}{2.303 \times 8.314}\left[\frac{320-300}{320 \times 300}\right] \end{aligned}

\begin{aligned} &\log 3=\frac{E_{a}}{2.303 \times 8.314}\left(\frac{20}{320 \times 300}\right)\\ &\frac{0.4771 \times 2.303 \times 8.314 \times 320 \times 300}{20}=E_{a}\\ &E_{a}=43848.486 \mathrm{J} / \mathrm{mol} \end{aligned}

 

Posted by

Sumit Saini

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 Board Result 2025 LIVE: Class 10th, 12th results by next week; DigiLocker codes, latest updates
anam-khan

Fake notice on CBSE 10th, 12th results 2025 in two-halves viral, board alerts students
anam-khan

CBSE Board Result 2025 LIVE: Class 10th, 12th results on cbse.nic.in soon; pass marks, DigiLocker code

Latest Question