By what factor is the rate of a reaction changed if an enzyme lowers the activation energy (Ea) by 7.0 kJ/mol at 37°C, according to the Arrhenius Equation?

• A. Approximately 2 times
• B. Approximately 5 times
• C. Approximately 10 times
• D. Approximately 20 times

Answer: (A)

Verified step by step guidance

1. Understand that the Arrhenius Equation is given by: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.
2. Convert the temperature from Celsius to Kelvin: \( T = 37 + 273.15 = 310.15 \text{ K} \).
3. Calculate the change in the exponent of the Arrhenius equation due to the change in activation energy: \( \Delta E_a = 7.0 \text{ kJ/mol} = 7000 \text{ J/mol} \).
4. Determine the factor by which the rate changes using the ratio of the rate constants: \( \frac{k_2}{k_1} = e^{\frac{\Delta E_a}{RT}} \). Substitute \( \Delta E_a = 7000 \text{ J/mol} \), \( R = 8.314 \text{ J/mol·K} \), and \( T = 310.15 \text{ K} \) into the equation.
5. Calculate the exponential term to find the factor by which the rate changes: \( e^{\frac{7000}{8.314 \times 310.15}} \). This will give you the approximate factor by which the rate of the reaction is increased.