Explanation:
The given data is as follows.
[tex]T_{1} = 4^{o}C[/tex] = (4 + 273) K = 277 K
[tex]T_{2} = 22^{o}C[/tex] = (22 + 273) K = 295 K
[tex]\frac{k_{2}}{k_{1}}[/tex] = 21.2
Formula to calculate activation energy will be as follows.
[tex]ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}][/tex]
[tex]ln (21.2) = \frac{E_{a}}{8.314 J/mol K} [\frac{1}{277} - \frac{1}{295}][/tex]
3.05 = [tex]\frac{E_{a}}{8.314 J/mol K} \times \frac{18}{81715}[/tex]
[tex]E_{a}[/tex] = 115116.914 J/mol
or = 115.116 kJ/mol
Thus, we can conclude that overall activation energy (kJ/mol) of the process is 115.116 kJ/mol.
