Correct Answer - C
According to Arrhenius equation
`k=Ae^(-E_(a)//RT)`
Taking natural logarithm of both sides, we get
`Ink=InA-(E_(a))/(RT)`
At temperature `T_(1))` , equation becomes
`Ink_(1)=InA-(E_(a))/(RT_(1)`
At temperature `T_(2))` , equation becomes
`Ink_(1)=InA-(E_(a))/(RT_(1)`
(Since `A` is constant for a given reaction). `K_(1)` and `k_(2)` are the value of rate constant at temperature `T_(1)` and `T_(2)` respectively.
Subtracting equation `(1)` from equation `(2)` , we obtain
`Ink_(2)-Ink_(1)=(-(E_(a))/(RT_(2)))-(-(E_(a))/(RT_(1)))`
`In(k_(2))/(k_(1))=-(E_(a))/(R)((1)/(T_(2))-(1)/(T_(1)))`
or log`(k_(2))/(k_(1))=-(E_(a))/(2.303R)((1)/(T_(2))-(1)/(T_(1)))`
or log`(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` Thus, we can evaluate the activation energy of a chemical reaction by knowing rate constants `(k_(1) and k_(2))` at two differences temperatures `(T_(1) and T_(2))` .