Chemistry 9th Edition

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

Chapter 12 - Chemical Kinetics - Review Questions: 10

Answer

$lnk=\frac{-E_a}{R}(\frac{1}{T})+lnA$ The data needed is the relationship between the rate constant and the temperature. In order to get a straight line, we would need to plot $lnk$ against $\frac{1}{T}$. The slope of the line would be $-\frac{E_a}{R}$, and the y-intercept would be $lnA$. The units of R are $\frac{J}{K\times mol}$. $E_a$ can be determined by using the following equation: $ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$.

Work Step by Step

Taking the natural log of both sides of the Arrhenius equation yields $lnk=ln(Ae^{-E_a/RT})$ which can be simplified to $lnk=lnA-\frac{E_a}{RT}$. Rearranging the terms gives $lnk=\frac{-E_a}{R}(\frac{1}{T})+lnA$. The relationship between the rate constant is needed in order to determine the equation. The equation is formatted in the form $y=mx+b$ where $y=lnk$, $m=\frac{-E_a}{R}$, $x=\frac{1}{T}$, and $b=lnA$, so to get a straight line, we must plot $y$ against $x$, or $lnk$ against $\frac{1}{T}$. The slope is equal to $m$ or $\frac{-E_a}{R}$. The R value is just the universal gas constant, or $8.3145\space\frac{J}{K\times mol}$, so its units are just $\frac{J}{K\times mol}$.
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.