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 the chemical reaction doubles for an increase of 10K in absolute temperature from 298K .Calculate Ea.

4.8     The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K Calculate E_{a}.

Answers (1)

best_answer

Given data 

T_{1}(initial temperature) = 298K and T_{2} (final temperature)= 308K

And we know that rate of reaction is nearly doubled when temperature rise 10-degree

So, \frac{k_{2}}{k_{1}}=2 and R = 8.314 J/mol/K

now, \log\frac{k_{2}}{k_{1}}=\frac{ E_{a}}{2.303}[\frac{T_{2}-T_{1}}{T_{1}T_{2}}]

On putting the value of given data we get,

  \log2=\frac{ E_{a}}{2.303}[\frac{10}{298\times 308}]

Activation energy (E_{a}) = \frac{2.303\times 8.314\times 298\times 308\times \log2}{10}Jmol^{-1}

                                      =52.9 KJ/mol(approx)

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question