Write down the Arrhenius equation relating the rate constant of reaction with temperature , mentioning what the terms indicate.
If `k_(1) and k_(2)` be the rate constant of a reaction at temperature `t_(1)^(@)C and t_(2)^(@)C` , respectively, find out the relation between `k_(1) ,k_(2) and t_(1) and t_(2)` . Given that the activation energy `(E_(a))` of the reaction remains unchanged within the temperature range mentioned.
The rate constant of a reaction at 400K and 500K are `0.02s^(-1) and 0.08s^(-1)` respectively . Determine the activation energy `(E_(a))` of the reaction.

Explanation

Second part : According to Arrhenius equation,
`k=Ae^(-(E_(a))/(RT))`
At temperature `t_(1)^(@)C and t_(2)^(@)C` , this equation takes the form,
`k_(1)=Ae ^(-(E_(a))/(R(273+t_(1))))and k_(2)^(-(E_(a))/(R(273+t_(2))))`
Dividing `k_(2)" by "k_(1) " gives"(k_(2))/(k_(1))=e.^((E_(a))/(R)[((t_(2)-t_(1)))/((273+t_(1))(273+t_(2)))])`
In`(k_(2))/(k_(1))=(E_(a))/(R)[((t_(2)-t_(1)))/((273+t_(1))(273+t_(2)))]`
`log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[((t_(2)-t_(1)))/((273+t_(1))(273+t_(2)))]`
This equation represents the relation between `k_(1),k_(2),t_(1) and t_(2)` .
Third part : To calculate activation energy `(E_(a))` , we use the relation , `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Putting the values of `T_(1) , T_(2) , k_(1) and k_(2)` into this equation ,we get `log.(0.08)/(0.02)=(E_(a))/(2.303xx8.314)((500-400)/(400xx500))`
or, `E_(a) = 23.05"kJ.mol"^(-1)`
Therefore , the activation energy of the reaction is `23.05"kJ.mol"^(-1)`.