Suppose that a catalyst lowers the activation barrier of a reaction from 122 kJ/mol to 54 kJ/mol. By what factor would you expect the reaction rate to increase at 25.00 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are the same.)

• A. Approximately 10^5 times
• B. Approximately 10^13 times
• C. Approximately 10^7 times
• D. Approximately 10^3 times

Answer: (B)

Verified step by step guidance

1. Understand that the rate of a chemical reaction is influenced by the activation energy, which is the energy barrier that must be overcome for reactants to transform into products. A catalyst lowers this barrier, increasing the reaction rate.
2. Use the Arrhenius equation to relate the rate constant \( k \) to the activation energy \( E_a \): \( k = A e^{-E_a/(RT)} \), where \( A \) is the frequency factor, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.
3. Convert the temperature from Celsius to Kelvin: \( T = 25.00 + 273.15 = 298.15 \) K.
4. Calculate the ratio of the rate constants for the catalyzed \( k_{cat} \) and uncatalyzed \( k_{uncat} \) reactions using the Arrhenius equation: \( \frac{k_{cat}}{k_{uncat}} = \frac{A e^{-E_{a,cat}/(RT)}}{A e^{-E_{a,uncat}/(RT)}} = e^{-(E_{a,cat} - E_{a,uncat})/(RT)} \).
5. Substitute the given activation energies into the equation: \( E_{a,uncat} = 122 \) kJ/mol and \( E_{a,cat} = 54 \) kJ/mol. Convert these values to J/mol (1 kJ = 1000 J) and calculate the factor by which the reaction rate increases.