Answer:
34.23 kJ/mol
Explanation:
So in this problem, we want to find out how much lower the activation energy is when an enzyme is present.
For same concentration level the reaction rate is determined by the rate factor, which is given by Arrhenius equation:
rate factor without catalyst:
k₁= A·e^{ - Ea₁ / (R·T) }
rate factor with catalyst:
k₂= A·e^{ - Ea₂ / (R·T) }
The ratio of the rate factors at same temperature is:
( k₂/k₁) = e^{ - Ea₂ / (R·T) } / e^{ - Ea₁ / (R·T) }
<=>
( k₂/k₁) = e^{ (Ea₁- Ea₂) / (R·T) }
But if we take the natural log of both sides and rearrange, we end up with Ea1 - Ea2. That's how much lower the activation energy will be.
Ea₁- Ea₂ = R·T·ln( k₂/k₁)
= 8.314472J/molk · 298K · ln(10^6)
= 34.23 kJ/mol