Correct Answer - A
According to the Arrhenius equation, we have.
`k=Ae^(-A_(a)//RT)`
Taking natural logarithm of both Siders:
In `k=InA-(E_(a))/(RT)`
Switching over to lograrithm to the base ten:
`2.303logk=2.303logA-(E_(a))/(RT)`
`{:(logk=,log A,-,(E_(a))/(2.303R),.(1)/T),(darr,darr,,darr,darr),(y=,c,+,m,x):}`
It corresponds to straigh line equation of the type, `y=c+mx` . Thyus, plot of log `kvs`//T` gives a straight line with intercept equal to log `A` and negative slope equal to `E_(a)//2.303R` . Thus, through the knowledge of slope, we can determine the energy of activation.