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By what factor does the rate constant change for a reaction when the temperature is increased from 115 °C to 230. °C if the reaction’s activation energy is 28.9 kJ/mol?

Respuesta :

sebassandin

Answer:

[tex]\frac{k_2}{k_1}=7.74[/tex]

Explanation:

Hello,

In this case, considering the Arrhenius law, the decreasing ratio is:

[tex]\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea}{RT_1} )}{Aexp(-\frac{Ea}{RT_2} )} \\[/tex]

By considering the 115°C as 1 and 230°C as 2, we obtain:

[tex]\frac{k_1}{k_2}=\frac{exp(-\frac{28900J/mol}{8.314J/molK*388.15K} )}{exp(-\frac{28900J/mol}{8.314J/molK*503.15K} )} =0.129[/tex]

And the increasing ratio:

[tex]\frac{k_2}{k_1}=\frac{1}{0.129}=7.74[/tex]

Best regards.

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