Question

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.)

Answer 1

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

Answer 2

The activation barrier with the presence of sucrose at the active site is lowered by the factor of 34.23 kJ/mol.

Activation energy is the amount of energy required by the reactant to convert into the product. The activation energy for the reaction is given as:

$k=A.e^{\frac{Ea_1}{RT} }$

Computation for the lowering of activation energy

The rate of reaction with the absence of sucrose at the active site is assumed to be 1.

The rate of reaction with the sucrose at the active site is $\rm 1-10^6$.

The rate of initial reaction is given as:

$k_1=A.e^\frac{Ea_1}{RT}$

The final rate of reaction is given as:

$k_2=A.e^\frac{Ea_2}{RT}$

The ratio of rate factor is given as:

$\dfrac{k_2}{k_2}=\dfrac{e^{Ea_1-Ea_2}}{RT}$

Substituting the values of R and T for the ratio of rate factor:

$ln \dfrac{k_2}{k_1}\;\times\;RT=Ea_1-Ea_2\\\\\text{ln} (10^-^6)\;\times\;8.314\;\text{J/mol.K}\;\times\;298\;\text K=Ea_1-Ea_2\\\\34.23 \;\text{kJ/mol}=Ea_1-Ea_2$

Thus, the final activation is lowered by the factor of 34.23 kJ/mol from the activation without sucrose.

Learn more about activation energy, here:

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