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07 KJ/mol(a) 50.43 KJ/mol(b) 34.37 KJ/mol(c) 82.31 KJ/molI 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

Answer»

Correct answer is (c) 82.31 KJ/mol

The explanation is: When we have TWO values of K and T

k1=koT1\(e^{-\FRAC{E}{RT1}} \) andk2=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 getln⁡\((\frac{k1}{k2})\)=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.


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