Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

The Arrhenius equation describes the relationship between the reaction rate constant \(k\) and activation energy \(E_a\), \(A\) is the pre-exponential factor, which is related to the collision factor, \(T\) is the temperature, and \(R\) is the gas constant: \(k = A e^{\frac{-E_a}{RT}}\)

\(\) For the uncatalyzed reaction, we have: \(k_{1} = A e^{\frac{-E_{a1}}{RT}}\) For the catalyzed reaction, we have: \(k_{2} = A e^{\frac{-E_{a2}}{RT}}\)

\(\) The given increase in reaction rate between the uncatalyzed and catalyzed reactions is \(1 \times 10^{5}\). So, the ratio of the reaction rate constants is: \(\frac{k_{2}}{k_{1}} = 1 \times 10^{5}\)

\(\) We can now substitute the expressions for the reaction rate constants from Step 2 into the equation obtained in Step 3: \(\frac{A e^{\frac{-E_{a2}}{RT}}}{A e^{\frac{-E_{a1}}{RT}}} = 1 \times 10^{5}\)

\(\) We can simplify the equation by cancelling out the pre-exponential factor \(A\) and taking the natural logarithm of both sides: \(\frac{e^{\frac{-E_{a2}}{RT}}}{e^{\frac{-E_{a1}}{RT}}} = 1 \times 10^{5}\) Taking the natural logarithm, we get: \(\frac{-E_{a2}}{RT} - \frac{-E_{a1}}{RT} = \ln{(1 \times 10^{5})}\) Now, we can solve for the difference in activation energy: \((E_{a1} -E_{a2}) = RT\ln{(1 \times 10^{5})}\) We are given a temperature of \(37^{\circ} \mathrm{C}\), which we must convert to Kelvin by adding 273.15: \(T = 37^{\circ} \mathrm{C} + 273.15 = 310.15 K \) Using \(R = 8.314 \mathrm{J \ mol^{-1} \ K^{-1}}\): \((E_{a1} -E_{a2}) = (310.15 K)(8.314 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}) \ln{(1 \times 10^{5})}\) Calculating the difference: \((E_{a1} -E_{a2}) = 16,120 \mathrm{J \ mol^{-1}} \approx 16.1 \mathrm{kJ \ mol^{-1}}\)

\(\) Thus, the enzyme must lower the activation energy by approximately 16.1 kJ/mol for a 100,000-fold increase in the reaction rate at physiological temperature.