Chemical Kinetics Question 20
Question: Activation energy is given by the formula [DCE 1999]
Options:
A) $ \log \frac{K_2}{K_1}=\frac{E_{a}}{2.303R}[ \frac{T_2-T_1}{T_1T_2} ] $
B) $ \log \frac{K_1}{K_2}=-\frac{E_{a}}{2.303R}[ \frac{T_2-T_1}{T_1T_2} ] $
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 modified form of Arrhenius equation.
