SATHEE: Chemical Kinetics Question 20

Chemical Kinetics Question 20

Question: Activation energy is given by the formula [DCE 1999]

Options:

A) $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

B) $\log \frac{K_{1}}{K_{2}} = - \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

C) $\log \frac{K_{1}}{K_{2}} = - \frac{E_{a}}{2.303 R} [\frac{T_{1} - T_{2}}{T_{1} T_{2}}]$

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

It is modified form of Arrhenius equation.

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