Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • The rate of reaction decreased by 5.555 times when the temperature was changed from <img alt="50^{\circ} \mathrm{C}" src="https://entrancecorner.oncodecogs.com/gif.latex?50%5E%7B%5Ccirc%7D%20%

The rate of reaction decreased by 5.555 times when the temperature was changed from 50^{\circ} \mathrm{C} to 35^{\circ} \mathrm{C}. The activation energy of the reaction will be \mathrm{\{ Take \ln 5.555=1.71\}}

Option: 1

\mathrm{39.84 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{94.29 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

\mathrm{24.69 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{44.6 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

The Arrhenices equation is 

\mathrm{K=A e{\frac{E a}{R T}}}

Assuming A and Ea to be independent of temperature

\mathrm{\begin{aligned} \ln \frac{K_2}{K_1} & =\frac{E a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\\\ \ln 5.555 & =\frac{E a}{8.314}\left(\frac{1}{308}-\frac{1}{323}\right) \\\\ \ln 5.555 & =\frac{E a}{8.514}\left(\frac{323-308}{99484}\right) \\\\ \ln 5.555 & =\frac{E a}{8.314}\left(\frac{15}{99484}\right) \\\\ 1.71 & =\frac{E a}{8.314}\left(\frac{15}{99484}\right) \end{aligned}}

\mathrm{\frac{9948 \times 1.71 \times 8.314}{15}=E a}

\mathrm{E a=94290.5=94.29 \mathrm{~kJ} /mol}

Posted by

Sumit Saini

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET 2026: Odisha aspirant dies by suicide fearing failure
anam-khan

NEET 2026 Answer Key (Out) LIVE: Challenge window link at neet.nta.nic.in soon; result date, cutoff
anam-khan

NEET UG Answer Key 2026 (OUT) LIVE: NTA provisional keys out neet.nta.ac.in; challenge window opens soon

Latest Question