Respuesta :
Answer : The activation energy for the reaction is, 43.4 KJ
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]K_1[/tex] = rate constant at [tex]18.0^oC[/tex]
[tex]K_2[/tex] = rate constant at [tex]37.0^oC[/tex] = [tex]3K_1[/tex]
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]18.0^oC=273+18.0=291K[/tex]
[tex]T_2[/tex] = final temperature = [tex]37.0^oC=273+37.0=310K[/tex]
Now put all the given values in this formula, we get:
[tex]\log (\frac{3K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{291K}-\frac{1}{310K}][/tex]
[tex]Ea=43374.66J/mole=43.4KJ[/tex]
Therefore, the activation energy for the reaction is, 43.4 KJ
The value of the activation barrier for the reaction will be "43.4 KJ". To understand the calculation, check below.
Arrhenius equation
According to the question,
At 18.0°C, rate constant = K₁
At 37.0°C, rate constant = K₂
Gas constant, R = 8.314 J/mole.K
Initial temperature, T₁ = 18.0°C or,
= 291 K
Final temperature, T₂ = 37.0°C or,
= 310 K
By using Arrhenius equation,
→ K = A × [tex]e^{\frac{-E_a}{RT} }[/tex]
or,
log([tex]\frac{K_2}{K_1}[/tex]) = [tex]\frac{E_a}{2.303\times R}[/tex] [[tex]\frac{1}{T_1}[/tex] - [tex]\frac{1}{T_2}[/tex]]
By substituting the values,
log([tex]\frac{3 K_1}{K_1}[/tex]) = [tex]\frac{E_a}{2.303\times 8.314}[/tex] [[tex]\frac{1}{291}[/tex] - [tex]\frac{1}{310}[/tex]]
[tex]E_a[/tex] = 43374.66 J/mole or,
= 43.4 KJ
Thus the above approach is correct.
Find out more information about Arrhenius equation here:
https://brainly.com/question/26757041
