Taking log on both sides, we get ln‌4‌k=ln‌A−‌
Ea2
RT
‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) On subtract Eq. (i) from Eq. (ii), we get ∴‌‌ln‌k−ln‌4‌k=‌
Ea2
RT
−‌
Ea1
RT
ln(‌
k
4k
)=‌
∆Ea
RT
‌−1.386=‌
⋮∆Ea
8.314×300
⇒‌‌∆Ea=−1.386×300×8.314 =−3456.96J∕mol Or ∆Ea=−3.45‌kJ‌mol−1 ∴ Change in activation energy of this reaction. =−3.456‌kJ∕mol‌. ‌