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

Options

A. `2.303xx2 cal`
B. `(2)/(2.303)cal`
C. `2 cal`
D. None of these

Answer

3

Explanation

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`