Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Rate constant   varies with temperature as given by equation: <img alt="\mathrm{\log \math

Rate constant \mathrm{k}  varies with temperature as given by equation:
\mathrm{\log \mathrm{k}\left(\mathrm{min}^{-1}\right)=5-2000 \mathrm{~K} / \mathrm{T}} . It can be said that -

A. pre-exponential factor \mathrm{A}  is 5
B. \mathrm{E_a}  is \mathrm{2000 \mathrm{kcal}}
C. pre-exponential factor \mathrm{\mathrm{A}}  is \mathrm{10^5} 
D. \mathrm{E_a}  is \mathrm{9.152 \mathrm{kcal}}

Option: 1

A only


Option: 2

C only


Option: 3

A and B only


Option: 4

C and D only


Answers (1)

best_answer

\mathrm{ \log \mathrm{k}=\log \mathrm{A}-\frac{E_{\mathrm{a}}}{2.303 R T} \\ }

\mathrm{ \text { Comparing it with } \log \mathrm{k}=5-\frac{2000}{T} \\ }

\mathrm{ \log _{10} \mathrm{~A}=5 \\ }
\mathrm{ \quad \Rightarrow \mathrm{A}=10^5 \\ }

\mathrm{\mathrm{E}_{\mathrm{a}} / 2.303 \mathrm{R}=2000 \\ }

\mathrm{\Rightarrow \mathrm{E}_{\mathrm{a}}=9.152 \mathrm{kcal} }.

 

Posted by

Devendra Khairwa

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 2026: IIT admissions show major shift toward AI, data science, away from core engineering courses
anam-khan

JEE Main 2026: NIT Delhi cut-offs for BTech courses show ever-increasing demands for AI, computer skills
anam-khan

597 Eklavya school students clear JEE, NEET exams in 2024-25; Gujarat, MP lead

Latest Question