Answer:
12.1 kJ
Explanation:
For this problem, we are going to use the Arrhenius Equation.
[tex]ln(\frac{k_{2} }{k_{1}})=\frac{E_{a} }{R} (\frac{1}{T_{1}} -\frac{1}{T_{2}} )\\ln(\frac{2}{1} )=\frac{E_{a} }{8.314} (\frac{1}{300}-\frac{1}{350} )[/tex]
We are going to solve for activation energy, [tex]E_{a}[/tex].
[tex]E_{a} =12.1[/tex] kJ