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.



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