Question

A second-order reaction was observed. The reaction rate constant at 25 oC was found to be 6.50 x 10-3L/mol and at 70 oC it was found to be 2.30 x 10-2 L/mol. Calculate the activation energy of this reaction.

Answer 1

Explanation:

Below is an attachment containing the solution.

Answer 2

The activation energy of the reaction is $23.86\;kJ$

Activation energy:

Activation energy is the energy required by the reactants in order to undergo any reaction to form products. If the reactants possess lesser energy than the activation energy then the reaction can not proceed further. The relation between activation energy, reaction rate, and temperature is described below:

Given that the reaction rate constants at temperatures $t_1=25^oC\;and\;t_2=70^oC\;are:\\\\k_1=6.50\times10^{-3}L/mol\;\;and\;k_2=2.30\times10^{-2}L/mol$

So the activation energy $E_a$ is given by:

$\ln\frac{k_2}{k_1} =\frac{E_a}{R}[\frac{1}{T_1}- \frac{1}{T_2} ]\\\\ \ln\frac{6.5\times10^{-3}}{2.3\times10^{-2}}=\frac{E_a}{R}[\frac{1}{298}-\frac{1}{342}]\\\\1.2639=E_a\times5.295\times10^{-5}\\\\E_a=23.86\;kJ$

Hence, the activation energy is $23.86\;kJ$

Learn more about activation energy:

https://brainly.com/question/11334504?referrer=searchResults