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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the rate constant for the reaction increases by a factor of 3.50 ✕ 103 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Answer :

ajeigbeibraheem

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

[tex]K=Ae^{\frac{-E}{RT}}[/tex]

If [tex]K_1=Ae^{\frac{-E_1}{RT}} -------equation 1[/tex]     &

[tex]K_2=Ae^{\frac{-E_2}{RT}} -------equation 2[/tex]

[tex]\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}[/tex]

[tex]E_1= E_2-RT*In(\frac{K_1}{K_2})[/tex]

[tex]E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})[/tex]

[tex]E_1 = 28957.39292[/tex]  [tex]J/mol[/tex]

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

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