Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

For a rxn \mathrm{a A+b B \rightarrow c C+d D,} the flot of \mathrm{\log \, K \, vs \frac{1}{T}} is given below. The temperature at which the rate constant of reaction is\mathrm{ 10^{-5} \mathrm{~s}^{-1}} is ___________  \mathrm{K}  \mathrm{{ Given \, \, that\, \, the\, \, rate \, \, constant \, \, of \, \, the \, \, reaction\, \, is 10^{-6} \mathrm{~s}^{-1} at \left.520 \mathrm{k}\right\}.}

Option: 1

614.7 k


Option: 2

713.6 k


Option: 3

538.67 k


Option: 4

413.87 k


Answers (1)

best_answer

\mathrm{\begin{aligned} & \text { According the Arrhenius equation, } \\\\ & \qquad \begin{aligned} \log K & =\log A-\frac{E a}{2.303 R T} \\\\ \text { Given Slope } & =\frac{E a}{2.303 R}=15,000 \\\\ K_1 & =10^{-5} \quad K_2=10^{-6} \\\\ T_1 & =520 K \\\\ \log 10\left[\frac{K_1}{K_2}\right] & =\frac{E a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\\\ \log \left[\frac{10^{-5}}{10^{-6}}\right] & =15000\left[\frac{1}{520}-\frac{1}{T_2}\right] \\\\ \log 10 & =15000\left[\frac{1}{520}-\frac{1}{T_2}\right] \\\\ T_2 & =538.67 K \end{aligned} \end{aligned}}

Posted by

Suraj Bhandari

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