Answer : The activation energy for the reaction is, 51.9 kJ
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at 295 K
[tex]K_2[/tex] = rate constant at 305 K = [tex]2K_1[/tex]
Ea = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 295 K
[tex]T_2[/tex] = final temperature = 305 K
Now put all the given values in this formula, we get:
[tex]\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}][/tex]
[tex]Ea=51879.96J=51.9kJ[/tex]
Therefore, the activation energy for the reaction is, 51.9 kJ
