Answer:
[tex]k_2=2.46[/tex]
Explanation:
Hello!
In this case, according to the Arrhenius equation for variable temperature:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Given the temperatures to be computed in kelvins and the activation energy, we obtain:
[tex]ln(\frac{k_2}{0.924} )=-\frac{89800J/mol}{8.314J/(mol*K)}(\frac{1}{306.45K} -\frac{1}{298.15K} )\\\\ln(\frac{k_2}{0.924} )=0.981\\\\k_2=0.924*exp(0.981)\\\\k_2=2.46[/tex]
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