07 KJ/mol A. 50.43 KJ/mol B. 34.37 KJ/mol C. 82.31 KJ/mol I had

07-kj-mol-a-50-43-kj-mol-b-34-37-kj-mol-c-82-31-kj-mol-i-had

07 KJ/mol

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

Correct answer is C. 82.31 KJ/mol

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.