TS EAMCET 18-July-2022 Shift 2 Solved Paper

According to Arrhenius theory,
k=Ae−‌

Ea

RT

Constant temperature, T=27∘C=300K
For initial reaction,
k=Ae(−‌

Eω1

RT

)
Taking log on both sides, we get
ln‌k=ln‌A−‌

Ea1

RT

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Forfinal reaction,
4k=Ae−‌

Ea2

RT

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. ‌