Suppose that a catalyst lowers the activation barrier of a reaction from 125 kJ/mol to 55 kJ/mol. By what factor would you expect the reaction rate to increase at 25 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

Verified step by step guidance

1. Identify the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$, where $k$ is the rate constant, $A$ is the frequency 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: $T = 25 + 273.15 = 298.15$ K.
3. Calculate the rate constant for the uncatalyzed reaction using the activation energy $E_{a1} = 125$ kJ/mol. Convert $E_{a1}$ to J/mol: $E_{a1} = 125,000$ J/mol.
4. Calculate the rate constant for the catalyzed reaction using the activation energy $E_{a2} = 55$ kJ/mol. Convert $E_{a2}$ to J/mol: $E_{a2} = 55,000$ J/mol.
5. Determine the factor by which the reaction rate increases by taking the ratio of the rate constants: $\frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a2}}{RT}}}{A e^{-\frac{E_{a1}}{RT}}} = e^{-\frac{E_{a2} - E_{a1}}{RT}}$.