The activation energy of a certain uncatalyzed biochemical reaction is 50.0 \(\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at $37^{\circ} \mathrm{C}\( the rate constant for the reaction increases by a factor of \)2.50 \times 10^{3}$ 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.

The Arrhenius equation is given by: $$ k = Ae^{\frac{-Ea}{RT}} $$ where: - k is the rate constant - A is the frequency factor - Ea is the activation energy - R is the gas constant, approximated to 8.314 J/(mol K) - T is the temperature in Kelvin We can write the Arrhenius equation for both the uncatalyzed and catalyzed reactions as follows, where E1 is the activation energy of the uncatalyzed reaction and E2 is the activation energy of the catalyzed reaction: Uncatalyzed reaction: $$ k_1 = Ae^{\frac{-E_1}{RT}} $$ Catalyzed reaction: $$ k_2 = Ae^{\frac{-E_2}{RT}} $$

Divide the Arrhenius equations for the catalyzed reaction (k2) by the uncatalyzed reaction (k1): $$ \frac{k_2}{k_1} = \frac{Ae^{\frac{-E_2}{RT}}}{Ae^{\frac{-E_1}{RT}}} $$ Since A is the same for both reactions, it cancels out: $$ \frac{k_2}{k_1} = \frac{e^{\frac{-E_2}{RT}}}{e^{\frac{-E_1}{RT}}} $$ Now, use the given information that the rate constant for the catalyzed reaction is 2.50 × 10^3 times greater than for the uncatalyzed reaction: $$ 2.50 \times 10^3 = \frac{e^{\frac{-E_2}{RT}}}{e^{\frac{-E_1}{RT}}} $$

Now, we need to simplify the equation further and isolate the activation energy of the catalyzed reaction (E2): $$ \frac{e^{\frac{-E_2}{RT}}}{e^{\frac{-E_1}{RT}}} = \frac{e^{\frac{-E_2-E_1}{RT}}}{} = 2.50 \times 10^3 $$ Now, take the natural logarithm (ln) of both sides: $$ \frac{-E_2-E_1}{RT} = ln(2.50 \times 10^3) $$ We are given the value of the activation energy for the uncatalyzed reaction, E1 = 50.0 kJ/mol, the temperature T = 37°C, and the gas constant R = 8.314 J/(mol·K). Convert the given temperature to Kelvin and the given activation energy to Joules: $$ T = 37+273.15 = 310.15\,K $$ $$ E_1= 50.0 \times 10^3\,J/mol $$ Now, substitute these values into the equation: $$ \frac{-E_2-50.0\times10^3}{8.314\times310.15} = ln(2.50 \times 10^3) $$ Now we can isolate E2 by first multiplying both sides by RT and then adding 50,000 to both sides: $$ E_2 = -RT\,ln(2.50 \times 10^3)+50.0 \times 10^3 $$

Substitute the values for R and T and solve for E2: $$ E_2 = -8.314 \times 310.15 \times ln(2.50 \times 10^3)+50.0 \times 10^3 $$ $$ E_2 \approx 38424\,J/mol $$ So, the activation energy for the catalyzed reaction is approximately 38.4 kJ/mol.