Chapter 14, Problem 38PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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.

â€‚Â ln(K1K2)â€Šâ€Š=â€Šâ€ŠEaR(1T2-1T1)T1=3Ã—102KT2â€Šâ€Šâ€Š=â€Šâ€Š3.18Ã—102KK1=KK2=3KÂ sinceÂ theÂ rateÂ constantÂ triplesÂ whenÂ temperatureÂ rises.

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

â€‚Â ln(K3K)â€Šâ€Š=â€Šâ€ŠEa8.314Jmolâˆ’1Kâˆ’1(13.18Ã—102K-13Ã—102K)ln(13)=â€Šâ€ŠEa8.314Jmolâˆ’1Kâˆ’1(1318K-1300K)lnÂ 0

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