At body temperature (37∘c) the rate constant of an enzyme-catalyzed reaction is 2.3 x 1014 times greater than the rate constant of the uncatalyzed reaction. assuming the frequency factor A is the same for both reaction, y how much does the enzyme lower the activation energ

The activation energy is lowered by +85.35 KJ/mol

Explanation:

Rate constant is related to the activation energy and the temperature of the reaction through

K = Ae⁻ᴱᵃ/ᴿᵀ

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

₁₂

At point 1, without the catalyst

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

Therefore, the activation energy is lowered by +85.35 KJ/mol

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