Respuesta :
Answer: [tex]0.58s^{-1}[/tex]
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]275K[/tex] = [tex]4.60\times 10^{-6}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]366K[/tex] = ?
[tex]Ea[/tex] = activation energy for the reaction = 108kJ/mol=108000 J/mol
R = gas constant = 8.314 J/moleK
[tex]T_1[/tex] = initial temperature = [tex]275K[/tex]
[tex]T_2[/tex] = final temperature = [tex]366K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{K_2}{4.60\times 10^{-6}})=\frac{108000}{2.303\times 8.314J/mole.K}[\frac{1}{275K}-\frac{1}{366K}][/tex]
[tex]K_2=0.58s^{-1}[/tex]
Therefore, the rate constant at 366 K is [tex]0.58s^{-1}[/tex]
The relation between the temperature and the rate of constant of a reaction.
The rate of constant at 366 K is 0.58 per second.
How to find the rate constant at particular temperature?
The relation between the temperature and the rate of constant of a reaction. It can be given as,
[tex]k=Ae\dfrac{-E\alpha}{RT}[/tex]
Here, [tex]k[/tex] is the rate of constant [tex]E[/tex] is the activation energy, [tex]R[/tex] is the gas constant (8.314 J/mol-K).
The above formula can also be given as,
[tex]\log \dfrac{k_2}{k_1}=\dfrac{Ea}{2.303R}\times[\dfrac{1}{T_1}-{\dfrac{1}{T_2} ]\\[/tex]
Given information-
The activation energy of the reaction is 108 kJ/mol.
The rate constant for the reaction is 4.60 x 10-6 s-1 at 275 K.
As the initial temperature is 275 K and the final temperature is 366 K. Thus put the values in the above formula as,
[tex]\log \dfrac{k_2}{4.60\times10^{-6}}=\dfrac{108000}{2.303\times8.314}\times[\dfrac{1}{275}-{\dfrac{1}{366} ]\\[/tex]
[tex]k_2=0.58\rm s^{-1}[/tex]
Thus the rate of constant at 366 K is 0.58 per second.
Learn more about the rate of constant here;
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