Chapter 14, Problem 37PS

### 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

# Calculate the activation energy, Ea, for the reaction2 N2O5(g) → 4 NO2(g) + O2(g)from the observed rate constants: k at 25 °C = 3.46 × 10−5 s−1 and k at 55 °C = 1.5 × 10−3 s−1.

Interpretation Introduction

Interpretation:

For the given reaction under given reaction conditions the activation energy for the 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

Given,

â€‚Â ln(K1K2)â€Šâ€Š=â€Šâ€ŠEaR(1T2-1T1)T1=â€Šâ€Š25âˆ˜Câ€Šâ€Š=â€Šâ€Š298.15KT2â€Šâ€Šâ€Š=â€Šâ€Š55âˆ˜C=328.15KK1=â€Šâ€Š3.46Ã—10-5sâˆ’1K2=Â 1.5Ã—10-3sâˆ’1Ea=â€Šâ€Š?

In order to determine the activation energy for the given reaction 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)Eaâ€Šâ€Š=â€Šâ€Šâ€Š

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