Answer:
Activation energy of the reaction is 79.5 kJ/mol
Explanation:
According to Arrhenius equation for a reaction-
[tex]k=Ae^{(\frac{-E_{a}}{RT})}[/tex]
where k is the rate constant, A is the Arrhenius constant, [tex]E_{a}[/tex] is the activation energy and T is temperature in kelvin
For the given two different set of condition, we can write-
at [tex]43^{0}\textrm{C}[/tex], [tex]5.22\times 10^{-4}=Ae^{(\frac{-E_{a}}{8.31\times 316})}[/tex]............(1)
at [tex]62^{0}\textrm{C}[/tex], [tex]2.91\times 10^{-3}=Ae^{(\frac{-E_{a}}{8.31\times 335})}[/tex]............(2)
[tex]Eq-(1)\div Eq-(2)[/tex] gives-
[tex]\frac{5.22\times 10^{-4}}{2.91\times 10^{-3}}=e^{\frac{E_{a}}{8.31}(\frac{1}{335}-\frac{1}{316})}[/tex]
Solving this equation we get [tex]E_{a}=79.5 kJ/mol[/tex]
So activation energy of the reaction is 79.5 kJ/mol
