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In Arthenius equation graph of log k vs ...

In Arthenius equation graph of `log k vs (1//T)` has slope equals to

A

`– E_a//2.303 R`

B

`E_a//R`

C

`E_a//2.303 R`

D

`- E_a//R`

Text Solution

Verified by Experts

The correct Answer is:
A
Hint : Arrhenius equation is `k = Ae^(-Ea//RT)`
or in `k = In A - E_a//RT`, so by plotting In `k vs 1//T`, Slope is `- E_a//R` and hence activation energy can be formed.
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ACCURATE PUBLICATION-CHEMICAL KINETICS-TOPIC : CHEMICAL KINETIC
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