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 relates the activation energy, rate constant, and temperature of a reaction: \(k = A e^{-E_a / RT}\) Where: - \(k\) is the rate constant - \(A\) is the frequency factor - \(E_a\) is the activation energy - \(R\) is the universal gas constant (\(8.314\ J/(mol\ K)\)) - \(T\) is the temperature in Kelvin

We are told that the rate constant for the catalyzed reaction is \(2.50 \times 10^{3}\) times greater than the rate constant for the uncatalyzed reaction. We are also given the activation energy for the uncatalyzed reaction and the temperature at which the reaction takes place. We will call the activation energy for the catalyzed reaction \(E_{ac}\). Given that the frequency factor is the same for both reactions, the ratio of their rate constants can be expressed as: \(\frac{k_{cat}}{k_{uncat}} = \frac{A e^{-E_{ac} / RT}}{A e^{-E_{a} / RT}}\) where: - \(k_{cat}\) is the rate constant for the catalyzed reaction - \(k_{uncat}\) is the rate constant for the uncatalyzed reaction Plugging in the given values: \(\frac{k_{cat}}{k_{uncat}} = 2.50 \times 10^{3}\) \(E_a = 50.0\ kJ/mol = 50,000\ J/mol\) \(T = 37^{\circ}C = 310K\) (converting to Kelvin: \(T = 37 + 273 = 310K\))

We have: \(\frac{A e^{-E_{ac} / RT}}{A e^{-E_a / RT}} = 2.50 \times 10^{3}\) Since the frequency factor \(A\) is the same for both reactions, it cancels out in our equation, giving: \(\frac{e^{-E_{ac} / RT}}{e^{-E_a / RT}} = 2.50 \times 10^{3}\) Let's isolate \(E_{ac}\) by multiplying both sides by \(e^{-E_a / RT}\) and taking the natural logarithm of both sides: \(-\frac{E_{ac}}{RT} = \ln{(2.50 \times 10^{3} e^{-E_a / RT})}\) Now, let's plug in the values of \(E_a\), \(R\), and \(T\): \(-\frac{E_{ac}}{8.314 \times 310} = \ln{(2.50 \times 10^{3} e^{-(50,000) / (8.314 \times 310)})}\) Next, multiply both sides by \(-8.314 \times 310\) to solve for \(E_{ac}\): \(E_{ac} = (8.314 \times 310) \ln{(2.50 \times 10^{3} e^{-(50,000) / (8.314 \times 310)})}\) Calculate the value of \(E_{ac}\) using a calculator: \(E_{ac} ≈ 33,825\ J/mol\) Convert the activation energy to kJ/mol (divide by 1000): \(E_{ac} ≈ 33.8\ kJ/mol\) So, the activation energy for the catalyzed reaction is approximately \(33.8\ kJ/mol\).