Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The rate constant for a first order reaction is given by the following equation : <img alt="\ln \mathrm{k}=33.24-\frac{2.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}" src="https://entrancecorne

The rate constant for a first order reaction is given by the following equation :
\ln \mathrm{k}=33.24-\frac{2.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}
The Activation energy for the reaction is given by___________ \mathrm{kJ} \mathrm{mol}^{-1} \cdot In Nearest integer)

(Given : \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} )

Option: 1

166


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\ln k=\ln A-\frac{E_{a}}{R T}.........(i)}

Given:

\mathrm{\operatorname{ln\; k}=33.24-\frac{2.0 \times 10^{4} K}{T}.....\text { (ii) }}

\mathrm{On~comparing \; Eq. (i)\; \& \; (ii), we ~have}

\mathrm{\frac{E a}{R}=2.0 \times 10^{4}}

\mathrm{Ea=2.0\times10^{4}\times8.3\; J/mol}

        \mathrm{=16.6\times10^{4}\; J/mol}

        \mathrm{=166\times10^{3}\; J/mol}

         \mathrm{=166\; KJ/mol}

Hence,the answer is 166.

Posted by

Sayak

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: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question