A hydrogenation reaction is carried out at 500K.If the same reaction is carried out in the presence of a catalyst at the same rate temperature required is 400K. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by `20"kJ.mol"^(-1)`

Explanation

From the Arrhenius equation , `logk=logA-(E_(a))/(2.030RT)`
`thereforelogk_(500)=logA-(E_(a_(1)))/(2.303RT_(1)),logk_(400)=logA-(E_(a_(2)))/(2.3030RT_(2))`
Given , `k_(500)=k_(400) thereforelogk_(500)=logk_(400)`
or, `(E_(a_(1)))/(T_(1))=(E_(a_(2)))/(T_(2)) " or, " (E_(a_(1)))/(500)=(E_(a_(2)))/(400 )" or, " (E_(a_(1)))/(E_(a_(2)))=(500)/(400)=(5)/(4)" " ...[1]`
Since the catalyst reduces the activation energy by `20"kJ.mol"^(-1),E_(a_(1))-E_(a_(2))=20`
Substituting `(E_(a_(1))-20) " for "E_(a_(2))` [1] , we get
`(E_(a_(1)))/(E_(a_(1))-20)=(5)/(4)`
or, `4E_(a_(1))=5E_(a_(1))-100 " or, "E_(a_(1))=100"kJ.mol"^(-1)`
`thereforeE_(a_(2))=100-20=80"kJ.mol"^(-1)`