Question

For a first order reaction a gives p the temperature t dependent rate constant k was found to follow the equation log ke bonus = - 2000 k upon t + 6the pre exponential factor and the activation energy ea respectively are

Answer 1

The activation energy is 38.3 kJmol−1

Explanation:

The logarithmic form of Arrhenius equation is

logk = log A−Ea 2.303 RT

Given, log k = 6 − 2000 T

Comparing the above two equations :

logA=6⇒A=106

and Ea 2.303 R=2000

⇒Ea=2000×2.303×8.314J

=38.3 kJmol−1

Thus the activation energy is 38.3 kJmol−1

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What is the activation energy for a reaction whose rate constant doubles when temperature changes from 30°C to 40°C?

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