Respuesta :
Answer : The activation energy of the reaction is, [tex]17.285\times 10^4kJ/mole[/tex]
Solution :
The relation between the rate constant the activation energy is,
[tex]\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = initial rate constant = [tex]4.55\times 10^{-5}L/mole\text{ s}[/tex]
[tex]K_2[/tex] = final rate constant = [tex]8.75\times 10^{-3}L/mole\text{ s}[/tex]
[tex]T_1[/tex] = initial temperature = [tex]195^oC=273+195=468K[/tex]
[tex]T_2[/tex] = final temperature = [tex]258^oC=273+258=531K[/tex]
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
[tex]\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}][/tex]
[tex]Ea=17.285\times 10^4kJ/mole[/tex]
Therefore, the activation energy of the reaction is, [tex]17.285\times 10^4kJ/mole[/tex]
Answer;
139.8 kJ/mol
Explanation;
The activation energy is given by the formula;
Eₐ = R•ln[k₂/k₁]•T₁T₂/(T₂ – T₁)
After converting ˚C to K:
T1 = 531 K
T2 = 468 k (after converting ˚C to K):
R = 8.314 J/mol•K
Eₐ = (8.314 J/mol•K) × ln[(3.2 × 10⁻³)/(4.5 × 10⁻⁵)] × (531 K)(468 K)/(531 – 468 K)
Eₐ = 139.8 kJ/mol
