Chemistry & Chemical Reactivity

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



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


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=rate constant

                    A=frequency factor

                    e=base of logarithms

                    Ea = energy of activation


Mathematically, the above equation can be written as,



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

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