Question

A certain bimolecular reaction at 40 °C at an activation energy of 30 kJ/mol. The addition of a catalyst reduces the activation energy by a factor of 2. How much faster does the catalyzed occur?

Select one:
OA. 318.63
OB. 358.63
C. 338.63
OD. 378.63

Answer 1

To find the rate enhancement due to the catalyst, we can use the Arrhenius equation:

k = Ae^(-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.

First, we need to convert the temperature from Celsius to Kelvin:

40°C + 273.15 = 313.15 K

Now, let's find the ratio of the rate constants for the catalyzed and uncatalyzed reactions:

k_catalyzed / k_uncatalyzed = e^((Ea_uncatalyzed - Ea_catalyzed) / RT)

Since the catalyst reduces the activation energy by a factor of 2:

Ea_catalyzed = 30 kJ/mol / 2 = 15 kJ/mol

Convert the activation energies to J/mol:

Ea_uncatalyzed = 30,000 J/mol
Ea_catalyzed = 15,000 J/mol

Now, plug in the values:

k_catalyzed / k_uncatalyzed = e^((30,000 J/mol - 15,000 J/mol) / (8.314 J/mol·K × 313.15 K))

k_catalyzed / k_uncatalyzed ≈ 358.63

The catalyzed reaction occurs approximately 358.63 times faster. So, the correct answer is OB. 358.63.