Respuesta :
Answer : The activation energy for the reaction is, 52.9 kJ/mol
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]T_1[/tex] = initial temperature = [tex]25.0^oC=273+25.0=298.0K[/tex]
[tex]T_2[/tex] = final temperature = [tex]25.0^oC+10=35.0^oC=273+35.0=308.0K[/tex]
[tex]K_1[/tex] = rate constant at [tex]25.0^oC[/tex] = [tex]12.5s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]35.0^oC[/tex] = [tex]2\times K_1=2\times 12.5s^{-1}=25.0s^{-1}[/tex]
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
Now put all the given values in this formula, we get:
[tex]\log (\frac{25.0s^{-1}}{12.5s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{298.0K}-\frac{1}{308.0K}][/tex]
[tex]Ea=52903.05J/mole=52.9kJ/mol[/tex]
Therefore, the activation energy for the reaction is, 52.9 kJ/mol
The activation energy for the reaction is 52.9 kJ/mol.
Activation Energy:
The minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport. It is given by Arrhenius equation which is as follows:
[tex]k=Ae^{\frac{-E_a}{RT}}[/tex]
Or, In terms of log it is given as:
[tex]log(\frac{k_2}{k_1})=\frac{E_a}{2.303*R} ( \frac{1}{T_1}-\frac{1}{T_2} )[/tex]
where,
T₁ , initial temperature = 280K
T₂ , final temperature = 308K
k₁ , rate constant = [tex]12.5s^{-1}[/tex]
k₂ , rate constant = [tex]25.0s^{-1}[/tex]
[tex]E_a[/tex] , activation energy for the reaction = ?
R, gas constant = 8.314 J/mol.K
On substituting the values we will get:
[tex]log(\frac{k_2}{k_1})=\frac{E_a}{2.303*R} ( \frac{1}{T_1}-\frac{1}{T_2} )\\\\log(\frac{25}{12.5})=\frac{E_a}{2.303*8.314} ( \frac{1}{298}-\frac{1}{308} )\\\\E_a=52903.05J/mol=52.9kJ/mol[/tex]
Thus, the activation energy for the reaction is 52.9 kJ/mol.
Find more information about Activation energy here:
brainly.com/question/1380484
