Activation energy of a chemical reaction can be determined by

Options

A. evaluting rate constant at standard temperature
B. evaluting velocities of reaction at two different temperatures
C. evaluating rate constants at two different temperatures
D. changing concentration of reactants

Answer

C

Explanation

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))` .