Question

A certain reaction has an activation energy of 51.64 kj/mol. At what temperature Kelvin will the reaction be 5.50 times faster than it did at 343K?

Answer 1

For this, we can use a manipulation of the Arrhenius equation...\[\ln( \frac{k_2}{k_1})=\frac{-E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})\]\[\ln( 5.50 )=\frac{-51640\frac{J}{mol}}{8.314 \frac{J}{K \cdot mol}}(\frac{1}{T_2}-\frac{1}{343K})\]Solving that for T_2, we get...\[T_2=378.646K=379K\]