Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Consider the two reaction having the rate constant as follow <img alt="\mathrm{A \longrightarrow B \quad K_{A}=10^{25} \mathrm{e}^{-6000 / T}}" src="https://entrancecorner.oncodecogs.com/gif

Consider the two reaction having the rate constant as follow
\mathrm{A \longrightarrow B \quad K_{A}=10^{25} \mathrm{e}^{-6000 / T}}
\mathrm{C \longrightarrow D \quad K_{C}=10^{20} \mathrm{e}^{-3000 / T}}
Determine the temperature at which rate constants of both the reaction are Equal.
 

Option: 1

200 \mathrm{~K}


Option: 2

250 \mathrm{~K}


Option: 3

300 \mathrm{~K}


Option: 4

260.5 \mathrm{~K}


Answers (1)

best_answer

Given that \mathrm{K_{A}=10^{25} e^{-6000 / T}: K_{C}=10^{20} e^{-3000 / T}}

Now, A/c to question

\mathrm{K_{A}=K_{c}}
\mathrm{10^{25} \mathrm{e}^{-6000 / T}=10^{20} \mathrm{e}^{-3000 / T}}
\mathrm{10^{5} e^{-6006 / T}=e^{-3000 / T}}
\mathrm{10^{5}=e^{\frac{6000-3000}{T}}=e^{+\frac{3000}{T}} }
\mathrm{\ln \left(10^{5}\right)=\frac{3000}{T} }
\mathrm{2.303 \times 5 \log (10)=\frac{3000}{T} }
\mathrm{T=\frac{3000}{2.303 \times 5}=260.5 \mathrm{~K} }.
\mathrm{T=260.5 \mathrm{~K}}

Posted by

himanshu.meshram

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 2026 Registration LIVE: NTA session 1 applications close on November 27; apply at jeemain.nta.nic.in
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

Latest Question