A. 50.43 KJ/mol
B. 34.37 KJ/mol
C. 82.31 KJ/mol
I had been asked this question in an international level competition.
I’m obligated to ask this question of Rate Laws and Stoichiometry Definitions topic in portion Rate Laws and Stoichiometry of Chemical Reaction Engineering
The explanation is: When we have two values of k and T
k1=koT1\(e^{-\frac{E}{RT1}} \) and k2=koT2\(e^{-\frac{E}{RT2}} \)
Modifying it gives
ln(k1)=ln(ko)+ln(T1) – \((\frac{E}{R})\frac{1}{T1}\) and ln(k2)=ln(ko)+ln(T2) – \((\frac{E}{R})\frac{1}{T2} \)
On further simplification we get ln\((\frac{k1}{k2})\)=ln\((\frac{T1}{T2}) – \frac{E}{R}(\frac{1}{T1}-\frac{1}{T2}) \)
ln\((\frac{0.002}{0.08})\)=ln\((\frac{273}{353})-\frac{E}{8.314}(\frac{1}{273} – \frac{1}{353}) \)
E = 34.377 KJ/mol.