Chemical Kinetics Question 273

Question: For a reaction, the rate constant is expressed as $ k=A{e^{-40000/T}} $ . The energy of the activation is

Options:

A) 40000 cal

B) 88000 cal

C) 80000 cal

D) 8000 cal

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $ k=A{e^{-E_{a}/RT}} $
$ \therefore \frac{-E_{a}}{R}=-40000 $
$ \therefore E_{a}=40000\times 2=80000,cal $



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