   # Would the slope of a ln( k ) versus 1/ T plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the ln( k ) versus 1/ T plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 22Q
Textbook Problem
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## Would the slope of a ln(k) versus 1/T plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the ln(k) versus 1/T plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

Interpretation Introduction

Interpretation: The rate laws for an uncatalyzed and a catalyzed reaction are given to be of the first-order overall. The comparison between the slope of a ln(k) versus 1T plot for a catalyzed and an uncatalyzed reaction is to be stated.

Concept introduction: The relationship between the rate constant and temperature is given by the Arrhenius equation, k=AeEaRT .

### Explanation of Solution

To determine: A comparison between the slope of a ln(k) versus 1T plot for a catalyzed and an uncatalyzed reaction.

The relationship between the rate constant and temperature is given by the Arrhenius equation. The Arrhenius equation is,

k=AeEaRT

Where,

• k is the rate constant.
• A is the Arrhenius parameter.
• Ea is the activation energy.
• R is the universal gas constant.
• T is the absolute temperature.

Take natural log on both sides in the above expression,

lnk=ln(AeEaRT)lnk=lnA+lneEaRTlnk=lnAEa<

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