# 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’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.