Question

The reaction rate of a bimolecular reaction at 300K is 10 times the reaction rate at 150K. Calculate the activation energy using collision theory.

(a) 4928 J/mol

(b) 5164 J/mol

(c) 3281 J/mol

(d) 1296 J/mol

I got this question during an interview for a job.

I'd like to ask this question from Rate Laws and Stoichiometry Definitions topic in section Rate Laws and Stoichiometry of Chemical Reaction Engineering

Answer 1

The correct answer is (a) 4928 J/mol

Easy explanation: T1 = 150K, T2 = 300K and k2 = 10k1

When we have two values of k and T

k1=ko\(\sqrt{T1}e^{-\frac{E}{RT1}}\) and  k2=ko\(\sqrt{T2}e^{-\frac{E}{RT2}}\)

Modifying it gives

ln⁡(k1)=ln⁡(ko)+0.5ln⁡(T1) – \((\frac{E}{R})\frac{1}{T1}\) and ln⁡(k2)=ln⁡(ko) – \((\frac{E}{R})\frac{1}{T2} \)

On further simplification we get  ln⁡⁡\((\frac{k1}{k2}) = 0.5ln⁡(\frac{T1}{T2}) \frac{E}{R} (\frac{1}{T1}-\frac{1}{T2}) \)

ln⁡\((\frac{k1}{10k1})\)=0.5ln⁡\((\frac{150}{300}) – \frac{E}{8.314}(\frac{1}{300} – \frac{1}{150}) \)

E = 4928 J/mol.