If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the active site of the enzyme? (assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and that the temperature is 298 k.)

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

answer; i believe the correct answer to the above question is ( friction)

we would need to see the picture to answer it