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 $
