Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 12 - Chemical Kinetics - Questions - Page 593: 22


The slope of a $ln(k)$ versus $\frac{1}{T}$ plot for a catalyzed reaction would be less negative.

Work Step by Step

The slope of a $ln(k)$ vs $\frac{1}{T}$ plot is equal to $\frac{-E_a}{R}$. A catalyzed reaction has a lower $E_a$ than an uncatalyzed one, and $R$ remains constant, so the slope should be less negative.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.