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Semenov [28]
3 years ago
9

La rapidez a la que cantan los grillos de árbol es de 2.0 ×102 veces por minuto a 27°C, pero es sólo de 39.6 veces por minuto a

5°C. A partir de estos datos, calcule la "energía de activación" para el proceso del canto.
Chemistry
1 answer:
Lorico [155]3 years ago
5 0

Answer:

51096J/mol = Energía de activación para el proceso del canto

Explanation:

Podemos solucionar este problema usando una forma de la ecuación de Arrhenius:

Ln\frac{K_2}{K_1} = \frac{-Ea}{R}  (\frac{1}{T_2} - \frac{1}{T_1} )

Donde K representa la velocidad de reacción (En este caso, la velocidad a la que cantan los grillos), Ea es la energía de activación, R la constante de los gases (8.314J/molK) y T representa la temperatura absoluta de 1, el estado inicial y 2, el estado final.

Estado inicial: K1 = 2.0x10²; T1 = 27°C + 273.15 = 300.15K

Estado final: K2 = 39.6; t2 = 5°C + 273.15 = 278.15K

Reemplazando en la ecuación:

Ln\frac{39.6}{2.0x10^2} = \frac{-Ea}{8.314J/molK}  (\frac{1}{278.15K} - \frac{1}{300.15K} )\\\\-1.6195 = \frac{-Ea}{8.314J/molK}*2.6351x10{-4}K{-1}

<h3>51096J/mol = Energía de activación para el proceso del canto</h3>
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