Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two Reactions and <img alt="\mathrm{P_2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

Two Reactions \mathrm{P_1} and \mathrm{P_2} have identical po exponential factors. The activation energy of \mathrm{P_1} is more than \mathrm{P_2} by \mathrm{30 \mathrm{kJmol}^{-1}}.If \mathrm{K_1} and \mathrm{K_2} are rate constants for reactions \mathrm{P_1} and \mathrm{P_2}, respectively at 300 k then \mathrm{\ln \left(\frac{k_2}{k_1}\right)} is equal to

Option: 1

8


Option: 2

6


Option: 3

4


Option: 4

12


Answers (1)

best_answer

By Arrhenius Equation 

\mathrm{\begin{aligned} k & =A e^{-E a_1 / R T} \\ k_1 & =A e^{-E a_1 / R T} \\ k_2 & =A e^{-E a_2 / R T} \\ \ln \left(\frac{k_1}{k_2}\right) & =e^{\left(E a_1-E a_2\right) / R T} \\ \ln \left(\frac{k_2}{k_1}\right) & =\frac{E a_1-E a_2}{R T}=\frac{30 \times 1000}{8.314 \times 300} \\ \ln \left(\frac{k_2}{k_1}\right) & =12 \end{aligned}}

Posted by

Ritika Jonwal

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 January 22 shift 1, 2 unofficial answer keys out; question paper solutions
anam-khan

JEE Main 2025 January 22 Shift 2 Exam Analysis: Theory-based questions from chemistry, maths rated lengthy
anam-khan

NTA moves JEE Main 2025 to January 28, 29 for 114 candidates citing technical snag at Karnataka centre

Latest Question