At body temperature (37°C), the rate constant of an enzymecatalyzed decomposition is 2.3X10¹⁴ times that of the uncatalyzed reac

Question

2 years ago

12

tion. If the frequency factor, A, is the same for both processes, by how much does the enzyme lower the Ea?

1 answer:

schepotkina [342] · Answer 1 · 2023-09-22T07:45:33+03:00

The activation energy is lowered by 85351.23 J/mol.

<h3>What is the activation energy?</h3>

The activation energy is the energy that is used to start a reaction.

In K = (In A) - (Ea / RT)

In K = (-Ea / RT) + In A

Where k = the activity constant

Ea = activation energy

R = molar gas constant

T = absolute temperature in Kelvin

In K₁ = (-Ea₁ / RT) + In A (eqn 1)

At point 2 with the catalyst

In K₂ = (-Ea₂ / RT) + In A (eqn 2)

Note that the molar gas constant, the absolute temperature in Kelvin, and the activity constant are all the same for both points.

Subtract (eqn 1) from (eqn 2)

In K₂ - In K₁ = (-Ea₂ / RT) - (-Ea₁ / RT) + In A - In A

In (K₂ / K₁) = (1/RT) (Ea₁ - Ea₂)

We were told that K₂ / K₁ = 2.3 × 10¹⁴, then find the difference in Ea

R = 8.314 J/mol.K, T = 37°C = 310.15 K

In (2.3 × 10¹⁴) = (1/ (8.314×310.15) (Ea₁ - Ea₂)

33.1 = (1/2578.5871) (Ea₁ - Ea₂)

(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol

Thus, the enzyme that lower the Ea is 85351.23 J/mol.

To learn more about activation energy, refer to the below link:

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