Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature 37 °C. In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kJ/mol must an enzyme lower the activation energy of the reaction to make it useful?

Verified step by step guidance

1. Step 1: Understand the Arrhenius equation, which relates the rate constant of a reaction to the activation energy and temperature: k = A * e^(-Ea/(RT)), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
2. Step 2: Convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature: T(K) = 37 + 273.15.
3. Step 3: Recognize that the reaction rate needs to be 5000 times faster with the enzyme, which implies that the catalyzed rate constant (k_cat) is 5000 times the uncatalyzed rate constant (k_uncat). Therefore, k_cat = 5000 * k_uncat.
4. Step 4: Use the Arrhenius equation for both the uncatalyzed and catalyzed reactions: k_uncat = A * e^(-Ea_uncat/(RT)) and k_cat = A * e^(-Ea_cat/(RT)). Set up the equation k_cat/k_uncat = 5000 = e^((Ea_uncat - Ea_cat)/(RT)).
5. Step 5: Solve for the difference in activation energy (Ea_uncat - Ea_cat) by taking the natural logarithm of both sides: ln(5000) = (Ea_uncat - Ea_cat)/(RT). Rearrange to find Ea_uncat - Ea_cat = R * T * ln(5000).