Chemical Kinetics Question 17
Activation energy is not given by a specific formula but is determined experimentally [DCE 1999]
Options:
A) $ \log \frac{K_2}{K_1}=\frac{E_{a}}{2.303R}\left[ \frac{T_2-T_1}{T_1T_2} \right] $
B) $ \log \frac{K_1}{K_2}=-\frac{E_{a}}{2.303R}\left( \frac{T_2-T_1}{T_1T_2} \right) $
C) $ \log \frac{K_1}{K_2}=-\frac{E_{a}}{2.303R}[ \frac{T_1-T_2}{T_1T_2} ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
It is a modified form of the Arrhenius equation.
BETA
