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Graph between log k and 1//T [k rate con...

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

A

`2.303xx2 cal`

B

`(2)/(2.303)cal`

C

`2 cal`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
3
We know, `logK=logA-(E_(a))/(2.303RT)`
compare this by `y=mx+c`
`m=-(E_(a))/(2.303R)`slpe of this
Given `-(E_(a))/(2.303R)=-(1)/(2.303)`
`E_(a)=R=2cal`
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