Answer:
The answer to the question is
The activation energy for the rotting reaction = 46.15 kJ/mol
Explanation:
To solve the question we have the Arrhenius's equation given by
k = [tex]Ae^{\frac{E_{a} }{RT} }[/tex]
Where Ea = Activation energy
R = Universal gas constant
T = temperature in kelvin
A = Arrhenius constant
k = Rate of reaction
k = 4 days to rot at 20 °C = [tex]Ae^{\frac{E_{a} }{R*293.15} }[/tex]
at 0 °C k = 16 days = [tex]Ae^{\frac{E_{a} }{R*273.15} }[/tex] therefore
k₂₀/k₀ = [tex]Ae^{\frac{E_{a} }{R*293.15} }[/tex] ÷ [tex]Ae^{\frac{E_{a} }{R*273.15} }[/tex] or
4/16 = [tex]e^{\frac{E_{a} }{R*293.15} }[/tex]÷[tex]e^{\frac{E_{a} }{R*273.15} }[/tex] or [tex]e^{\frac{E_{a} }{2270.9691} }[/tex] = 4 ×[tex]e^{\frac{E_{a} }{2437.2491} }[/tex]
[tex]{\frac{E_{a} }{2270.9691} }[/tex] = ln 4 + [tex]{\frac{E_{a} }{2437.2491} }[/tex] or Ea = 46145.21 J/mol or 46.15 kJ/mol
