07 KJ/mol

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
Correct answer is C. 82.31 KJ/mol

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.