Two reactions R1 and R2 have identical pre-exponential factors.  Activation energy of R1 exceeds that of R2 by 10 kJ mol−1.  If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R=8.314 J mol−1K−1)  
Option: 1 6
Option: 2 4
Option: 3 8
Option: 4 12
 

Answers (1)

K=Ae^{\frac{-Ea}{RT}}

\frac{K_{1}}{K_{2}}=e\left ( \frac{Ea_{1}-Ea_{2}}{RT} \right )

Taking log both the sides

ln\left ( \frac{K_{2}}{K_{1}} \right )=\frac{E_{a1}-E_{a2}}{RT}= \frac{10000}{8.314\times 300}=4

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, Faculty Support.

₹ 7999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions