Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The rate of doubles when its temperature changes from 290 K to 300 K. Activation e

The rate of \mathrm{r_{xn}} doubles when its temperature changes from 290 K to 300 K. Activation energy of such reaction will be log [2=0.301]

Option: 1

32197


Option: 2

50140


Option: 3

24316


Option: 4

27819


Answers (1)

best_answer

\mathrm{\begin{aligned} \log \frac{k_2}{k_1} & =\frac{-E a}{2.303 R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \\ \log \frac{k_2}{k_1} & =\frac{E a}{2.30 B R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ \log 2 & =\frac{E a}{2.303 \times 8.314}\left(\frac{30-2 a}{8700}\right) \\ \log 2 & =\frac{E a}{2.303 \times 8.314}\left(\frac{1}{8700}\right) \\ E a & =\log 2 \times 2.303 \times 8.314 \times 8700 \\ E a & =0.301 \times 2.303 \times 8.314 \times 8700 \\ E a & =50140 \end{aligned}}

Posted by

sudhir.kumar

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 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question