Answer:
a) [tex]Ea=-131.522x10^{3}\frac{J}{mol}[/tex]
b) [tex]k_3=1.98x10^{-5}s^{-1}[/tex]
Explanation:
Hello,
a) In this case, for this kinetics problem, we can consider the temperature-dependent Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )[/tex]
In such a way, for the given temperatures and rate constant, we compute the activation energy as follows:
[tex]ln(\frac{0.228s^{-1}}{0.000122s^{-1}} )=\frac{Ea}{R}(\frac{1}{(77+273)K}-\frac{1}{(27+273)K} )\\\\7.533=\frac{Ea}{R}*-4.762x10^{-4}K^{-1}\\ \\Ea=R\frac{7.533}{-4.762x10^{-4}K^{-1}} \\[/tex]
[tex]Ea=8.314\frac{J}{mol*K}*-15819.3K\\\\Ea=-131.522x10^{3}\frac{J}{mol}[/tex]
b) In this case, we use the previously computed activation energy in order to compute the rate constant at the asked 17°C:
[tex]k_3=k_1exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))\\\\k_3=0.000122s^{-1}exp[\frac{-131.522x10^3\frac{J}{mol} }{8.314\frac{J}{mol*K}}*(\frac{1}{(17+273)K} -\frac{1}{(27+273)K} )]\\\\k_3=1.98x10^{-5}s^{-1}[/tex]
Best regards.
