Respuesta :
Answer:
Explanation:
We shall apply Arrhenius equation which is given below .
[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]
K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .
Putting the given values
[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]
[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]
T₂ = 335.27 K
= 62.27 °C
The higher temperature is 62.27°C.
Calculating the higher temperature:
Given that the activation energy of the reaction is:
Eₐ = 39.5 kJ/mol
initial temperature T₁ = 25°C = 298K
Let the final temperature be T₂
The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.
According to the question: K₂/K₁ = 5.9
Now, applying the Arrhenius equation we get:
[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]
T₂ = 335.27K
T₂ = 335.27 -273
T₂ = 62.27°C
Learn more about rate constant:
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