Question

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?

Answer 1

Answer:

$\frac{k_2}{k_1}=7.74$

Explanation:

Hello,

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

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea}{RT_1} )}{Aexp(-\frac{Ea}{RT_2} )}$

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

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

And the increasing ratio:

$\frac{k_2}{k_1}=\frac{1}{0.129}=7.74$

Best regards.