To solve the problem, we will follow these steps:
### Step 1: Determine the Activation Energies
The activation energy for the uncatalyzed reaction (Ea1) is given as:
\[
Ea1 = 100 \, \text{kJ/mol}
\]
The presence of a catalyst lowers the activation energy by 75%. Therefore, the activation energy for the catalyzed reaction (Ea2) can be calculated as:
\[
Ea2 = Ea1 - 0.75 \times Ea1 = 100 - 75 = 25 \, \text{kJ/mol}
\]
### Step 2: Convert Activation Energies to Joules
Since the gas constant \( R \) is typically expressed in J/(mol·K), we need to convert the activation energies from kJ/mol to J/mol:
\[
Ea1 = 100 \, \text{kJ/mol} = 100,000 \, \text{J/mol}
\]
\[
Ea2 = 25 \, \text{kJ/mol} = 25,000 \, \text{J/mol}
\]
### Step 3: Use the Arrhenius Equation
The Arrhenius equation relates the rate constants \( k_1 \) and \( k_2 \) for the uncatalyzed and catalyzed reactions, respectively:
\[
k_1 = A e^{-\frac{Ea1}{RT}}
\]
\[
k_2 = A e^{-\frac{Ea2}{RT}}
\]
Dividing the two equations gives:
\[
\frac{k_2}{k_1} = \frac{A e^{-\frac{Ea2}{RT}}}{A e^{-\frac{Ea1}{RT}}} = e^{-\frac{Ea2 - Ea1}{RT}}
\]
### Step 4: Substitute the Values
Now we substitute \( Ea1 \) and \( Ea2 \):
\[
\frac{k_2}{k_1} = e^{-\frac{25,000 - 100,000}{RT}} = e^{-\frac{-75,000}{RT}} = e^{\frac{75,000}{RT}}
\]
### Step 5: Take the Logarithm
Taking the logarithm (base 10) of both sides:
\[
\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{75,000}{2.303 \cdot RT}
\]
### Step 6: Substitute R and T
Using \( R = 8.314 \, \text{J/(mol·K)} \) and \( T = 27^{\circ}C = 300 \, K \):
\[
\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{75,000}{2.303 \cdot 8.314 \cdot 300}
\]
### Step 7: Calculate the Value
Calculating the denominator:
\[
2.303 \cdot 8.314 \cdot 300 = 19.147
\]
Now substituting this value:
\[
\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{75,000}{19.147}
\]
Calculating this gives:
\[
\log_{10} \left( \frac{k_2}{k_1} \right) \approx 3,911.5
\]
### Final Answer
Thus, the value of \( \log_{10} \left( \frac{k_2}{k_1} \right) \) is approximately:
\[
\log_{10} \left( \frac{k_2}{k_1} \right) \approx 3.9115
\]
