A reaction proceeds with ∆ H = 40 kJ/mol. The energy of activation of the uncatalyzed reaction is 120 kJ/mol, whereas i t is 55 kJ/mol for the catalyzed reaction. How many times faster is the catalyzed reaction than the uncatalyzed reaction at 25°C?
To calculate how many times faster the catalyzed reaction is than the uncatalyzed reaction, we need to use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy and temperature.
The Arrhenius equation is given by:
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))
- T is the temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, let's calculate the rate constants for the uncatalyzed and catalyzed reactions using the Arrhenius equation.
For the uncatalyzed reaction:
k_uncatalyzed = A * e^(-Ea_uncatalyzed/RT)
k_uncatalyzed = e^(-120,000 J/mol / (8.314 J/(mol*K) * 298.15 K))
For the catalyzed reaction:
k_catalyzed = A * e^(-Ea_catalyzed/RT)
k_catalyzed = e^(-55,000 J/mol / (8.314 J/(mol*K) * 298.15 K))
Now, we can calculate the ratio of the rate constants to determine how many times faster the catalyzed reaction is compared to the uncatalyzed reaction:
times_faster = k_catalyzed / k_uncatalyzed
Substituting the values we calculated for the rate constants:
times_faster = (e^(-55,000 J/mol / (8.314 J/(mol*K) * 298.15 K))) / (e^(-120,000 J/mol / (8.314 J/(mol*K) * 298.15 K)))
Evaluating this expression will give us the desired result.