1.

Arrhenius equation to determine the activation energy.(a) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}) \)(b) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}+\frac{1}{T1}) \)(c) ln⁡\((\frac{k1}{k2}) = -\frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}) \)(d) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{T1}{T2}) \)I had been asked this question during an internship interview.My question is based upon Rate Laws and Stoichiometry Definitions in section Rate Laws and Stoichiometry of Chemical Reaction Engineering

Answer»

Correct OPTION is (a) ln⁡\((\frac{K1}{k2}) = \frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}) \)

To elaborate: When we have two values of k and T

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

Modifying it gives

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

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


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