[A]:The Arrhenius equation k=Ae^(-E_(a)//RT) gives the relation between rate constant and temperature. [R]:The graph of log kto(1)/(T) is linear and with the help of this calculation of energy of activation can be possible.

[A]:The Arrhenius equation k=Ae^(-E_(a)//RT) gives the relation betwee

1.	[A]:The Arrhenius equation k=Ae^(-E_(a)//RT) gives the relation between rate constant and temperature. [R]:The graph of log kto(1)/(T) is linear and with the help of this calculation of energy of activation can be possible.
Answer» ASSERTION [A] and reason [R] both are correct and [R] gives correct EXPLANATION of [A] Assertion [A] and reason [R] both are correct but [R] does not give correct explanation of [A] Assertion [A] is wrong but Reason [R] is wrong Assertion [A] is wrong but Reason [R]is correct Solution :By taking log both side of Atthenious EQUATION, In k=In A`-(E_(a))/(RT)` and log k=log A`-(E_(a))/(2.303R)((1)/(T))` The equation is obtain.This euqation is equal to y=mc+c.So,in graph log `kto(1)/(T)`,the valie of slope is `(E_(a))/(2.303R)` So,the value of `E_(a)` is CALCULATED by getting of slope and the explanation of Arrhenius equation is obtain from graph.