Answer:
[tex]Ea=-84.9\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, temperature-variable version of the Arrhenius equation is applied in order to compute the activation energy as shown below:
[tex]ln(\frac{k_2}{k_1})=\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} ) \\\\Ea=R*\frac{ln(\frac{k_2}{k_1})}{(\frac{1}{T_2} -\frac{1}{T_1} )}[/tex]
Thus, we replace each variable considering k2=3*k1 (tripling):
[tex]Ea=8.314\frac{J}{mol*K} *\frac{ln(\frac{3k_1}{k_1})}{(\frac{1}{310K} -\frac{1}{300K} )}\\\\Ea=-84.9\frac{kJ}{mol}[/tex]
Best regards.
