Answer:
[tex]k_2=0.504s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:
[tex]ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}[/tex]
Regards!
