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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:
c
`2.303log_(10)K=-E_(a)/(RT)+2.303log A'` is Arrhenius
equation `(K=A' e^(-E_(a)//RT))`
Thus, `-E_(a)/(2.303R)=tantheta=-1/2.303`
`:. E_(a)=R=2 cal`
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