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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

When heated, cyclopropane is converted to propene (Example 14.5) Rate constants for this reaction at 470 °C and 510 °C are k = 1.10 × 10−4 s−1 and k = 1.02 × 10−3 s−1, respectively. Determine the activation energy, Ea, from these data.

Interpretation Introduction

Interpretation:

For the given reaction under given reaction conditions that is given rate constant, temperature the activation energy for the reaction should be determined.

Concept introduction:

In order to establish the plausibility of a mechanism, one must compare the rate law of the rate determining step to the experimentally determined rate law.

Rate determining step: In a chemical reaction the rate determining step is the slowest step in which the rate of the reaction depends on the rate of that slowest step.

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.

Intermediate species: It is the species formed during the middle of the chemical reaction between the reactant and the desired product.

Arrhenius equation:

  • Arrhenius equation is a formula that represents the temperature dependence of reaction rates
  • The Arrhenius equation has to be represented as follows

  k=AeEa/RTlnk=lnAeEa/RTlnk=(EaR)(1T)+lnA

  • Ea represents the activation energy and it’s unit is kJ/mol
  • R represents the universal gas constant and it has the value of 8.314 J/K.mol
  • T represents the absolute temperature
  • A represents the frequency factor or collision frequency
  • e represents the base of natural logarithm
  •  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
Explanation

Given:

  ln(K1K2)=EaR(1T2-1T1)T1=470CT2=510CK1=1.10×10-4s1K2=1.02×10-3s1 Ea=?

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.

  ln(K1K2)=EaR(1T2-

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