The rate constant for a reaction is given by the Arrhenius equation:
\[ k = A \cdot e^{-\frac{E_a}{RT}} \]
Where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor (frequency factor)
- \( E_a \) is the activation energy
- \( R \) is the gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin
Given that the activation energy is lowered by 25 kJ/mol and we want to compare the rate with and without the catalyst, we can compare the rate constants for the two cases.
Let's assume the initial rate constant (without the catalyst) is \( k_1 \) and the rate constant with the catalyst is \( k_2 \).
For the initial rate (without the catalyst):
\[ k_1 = A \cdot e^{-\frac{E_{a1}}{RT}} \]
For the rate with the catalyst (lowered activation energy):
\[ k_2 = A \cdot e^{-\frac{E_{a2}}{RT}} \]
Given that \( E_{a2} = E_{a1} - 25 \) kJ/mol, we can substitute this into the equation for \( k_2 \):
\[ k_2 = A \cdot e^{-\frac{(E_{a1} - 25)}{RT}} \]
To find how many times the rate grows, we can take the ratio of \( k_2 \) to \( k_1 \):
\[ \frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{(E_{a1} - 25)}{RT}}}{A \cdot e^{-\frac{E_{a1}}{RT}}} = e^{-\frac{25}{RT}} \]
Now, plug in the temperature \( T = 25°C = 298.15 \, \text{K} \):
\[ \frac{k_2}{k_1} = e^{-\frac{25}{298.15 \cdot 8.314}} \approx 24069 \]
So, the rate grows approximately 24069 times when the activation energy is lowered by 25 kJ/mol at 25°C by adding a catalyst. |
