Answer: The activation energy of the reaction is 124.6 kJ/mol
Explanation:
To calculate activation energy of the reaction, we use Arrhenius equation, which is:
[tex]\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_{79^oC}[/tex] = equilibrium constant at 79°C = [tex]0.394s^{-1}[/tex]
[tex]K_{26^oC}[/tex] = equilibrium constant at 26°C = [tex]2.08\times 10^{-4}s^{-1}[/tex]
[tex]E_a[/tex] = Activation energy of the reaction = ?
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]26^oC=[26+273]K=299K[/tex]
[tex]T_2[/tex] = final temperature = [tex]79^oC=[79+273]K=352K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol[/tex]
Hence, the activation energy of the reaction is 124.6 kJ/mol
