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Calculate the activation energy, E a Ea , in kilojoules per mole for a reaction at 57.0 ∘ C 57.0 ∘C that has a rate constant of 0.235 s − 1 0.235 s−1 and a frequency factor of 3.41 × 10 11 s − 1 3.41×1011 s−1 .

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Explanation:

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The activation energy will be 76.604kJ.

Relation between Activation energy and Rate constant:

[tex]k=A^{(-E_a/RT)}[/tex]

where

  • K= rate constant
  • Ea=  Activation energy
  • T= temperature
  • A= Arrhenius constant

Given:

[tex]A= 3.41 * 10^{11} s^{- 1}[/tex]

[tex]k = 0.235 s^{-1}[/tex]

[tex]T= 57^oC[/tex] = 57+273= 330K

To find:

Ea=?

On substituting the values in the above formula:

[tex]k=A^{(-E_a/RT)}\\\\k=A exp (-E_a/RT)\\\\\frac{0.235}{3.4*10^{11}} =exp (\frac{-E_a}{8.314*330})\\\\ ln (\frac{0.235}{3.14*10^{11}}) = \frac{-E_a}{8.314*330}\\\\E_a=76604.14J\\\\E_a=76.604kJ[/tex]

Thus, the activation energy will be 76.604kJ.

Find more information about Activation energy here:

brainly.com/question/9950855

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