   Chapter 14, Problem 38PS

Chapter
Section
Textbook Problem

If the rate constant for a reaction triples when the temperature rises from 3.00 × 102 K to 3.10 × 102 K, what is the activation energy of the reaction?

Interpretation Introduction

Interpretation:

For the reaction with given condition that rate constant of the reaction triples under given temperature range hence the activation energy for that reaction should be determined.

Concept Introduction:

Arrhenius equation:

Arrhenius equation is used to calculate the rate constant of many reactions. Arrhenius equation takes the form

k=Ae-Ea/RT

Where,

k=rate constant

A=frequency factor

e=base of logarithms

Ea = energy of activation

T=Temperature

Mathematically, the above equation can be written as,

lnk=lnA-EaRT

Explanation

In order to find the activation energy we need to use the following expression which relates the rate constant, activation energy and the temperature.

ln(K1K2)=EaR(1T2-1T1)T1=3×102KT2=3.18×102KK1=KK2=3K since the rate constant triples when temperature rises.

With known rate constant and the temperature the activation energy is calculated as follows,

ln(K3K)=Ea8.314Jmol1K1(13.18×102K-13×102K)ln(13)=Ea8.314Jmol1K1(1318K-1300K)ln 0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 