Respuesta :
Explanation:
Relation between Arrhenius energy and rate constant is as follows.
[tex]ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}][/tex]
The given data is as follows.
[tex]\frac{k_{2}}{k_{1}}[/tex] = 3 times.
R = 8.314 J/mol K
[tex]T_{1} = 20^{o} + 273 K[/tex]
= 293 K
[tex]T_{1} = 40^{o} + 273 K[/tex]
= 313 K
Now, putting the given values into the above formula as follows.
[tex]ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}][/tex]
[tex]ln (3) = \frac{E_{a}}{8.314 J/mol K} [\frac{1}{293 K} - \frac{1}{313 K}][/tex]
1.098 = [tex]\frac{E_{a}}{8.314 J/mol K} \times 2.18 \times 10^{-4}[/tex]
[tex]E_{a} = 4.187 \times 10^{4}[/tex]
= 41870 J/mol
Thus, we can conclude that Arrhenius energy of activation for the reaction is 41870 J/mol.
The required value of Arrhenius energy of activation for the given reaction is [tex]4.187 \times 10^{4} \;\rm J/mol[/tex].
Given data:
The initial temperature of reaction is, [tex]T = 40.0^{\circ} \:\rm C[/tex].
The final temperature of the reaction is, T' = 20.0° C.
The rate constant at initial temperature is 3 times that of final temperature.
The given problem is based on the relation between the Arrhenius energy and the rate constant. And the mathematical expression for the same is given as,
[tex]ln \dfrac{k'}{k}=\dfrac{E_{a}}{R} \times [ \dfrac{1}{T}-\dfrac{1}{T'}][/tex]
Here, R is the universal gas constant. And [tex]E_{a}[/tex] is the Arrhenius energy of activation.
As given in the problem,
[tex]\dfrac{k'}{k}=3[/tex]
Solving as,
[tex]ln \; 3=\dfrac{E_{a}}{8.314} \times [ \dfrac{1}{(20+273)}-\dfrac{1}{(40+273)}]\\\\\\1.098=\dfrac{E_{a}}{8.314} \times (2.18 \times 10^{-4})\\\\\\E_{a}=4.187 \times 10^{4} \;\rm J/mol[/tex]
Thus, we can conclude that the required value of Arrhenius energy of activation for the given reaction is [tex]4.187 \times 10^{4} \;\rm J/mol[/tex].
Learn more about the Arrhenius Energy here:
https://brainly.com/question/16296189
