Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  Catalyst R reduces the activation energy for a reaction by <img alt="\mathrm{5 \mathrm{~K} \mathrm{~J} \mathrm{~mol}^{-1} \, at\, 250 \mathrm{~K}}" src="https://entrancecorner.oncodecogs.com/

 Catalyst R reduces the activation energy for a reaction by \mathrm{5 \mathrm{~K} \mathrm{~J} \mathrm{~mol}^{-1} \, at\, 250 \mathrm{~K}}. The ratio of rate constants,

\mathrm{\frac{K T\text {, catalysed }}{K T \text {, uncatalysed }} \, \, is \, \, e^x}.Calculate the value of \mathrm{x}

Option: 1

e^{3.6}


Option: 2

e^{4.7}


Option: 3

e^{5.3}


Option: 4

e^{2.4}


Answers (1)

best_answer

By Arrhenius Eq:-
\mathrm{\begin{aligned} & K=A e^{\frac{-E a}{R T}} \\ & \text { Kcatalyst }=A e^{\frac{E a}{R T}} \\ & \text { Kuncatalyst }=A e^{\frac{-E a}{R T}} \\ & \frac{\text { Kcatalyst }}{\text { Kuncatalyst }}=e^{\frac{E a-E^{\prime} a}{R T}}=e^{\frac{5 \times 1000}{8.314 \times 250}}=e^{2.4} \end{aligned} }

Posted by

Suraj Bhandari

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: Previous year NIT Calicut cut offs; opening, closing ranks
anam-khan

JEE Main 2025 Admit Card Live: Session 1 hall ticket soon at jeemain.nta.nic.in; photo correction, cut-offs
anam-khan

NTA finds errors in JEE Main 2025 photos; asks to upload correct images by January 17

Latest Question