The energy of activation for ab uncatalysed reaction is `100KJ mol ^(-1)` presence of a catalyst lowers the energy of activation by 75% . The `log _(10).K_2/K_1` of the ratio of rate constant of catalysed and uncatalysed reactions at `27^@C` is ?
Assume the frequency factor is same for both reactions. ( Given `2.303xx8.314 = 19.147 `)

Options

A. 13.05
B. `26.10`
C. `6.52`
D. None of these

Answer

To solve the problem, we need to find the logarithm of the ratio of the rate constants of the catalyzed and uncatalyzed reactions. We will use the Arrhenius equation and the information given in the problem.

### Step-by-Step Solution:

1. **Identify the Activation Energies**:
- The activation energy for the uncatalyzed reaction (EA1) is given as 100 kJ/mol.
- The catalyst lowers the activation energy by 75%. Therefore, the activation energy for the catalyzed reaction (EA2) is:
\[ EA2 = EA1 - (0.75 \times EA1) = 100 \, \text{kJ/mol} - 75 \, \text{kJ/mol} = 25 \, \text{kJ/mol} \]

2. **Convert Activation Energies to Joules**:
- Convert kJ/mol to J/mol for both activation energies:
\[ EA1 = 100 \, \text{kJ/mol} = 100,000 \, \text{J/mol} \]
\[ EA2 = 25 \, \text{kJ/mol} = 25,000 \, \text{J/mol} \]

3. **Use the Arrhenius Equation**:
- The Arrhenius equation states:
\[ k = A e^{-\frac{EA}{RT}} \]
- For the uncatalyzed reaction (K1):
\[ K1 = A e^{-\frac{EA1}{RT}} = A e^{-\frac{100,000}{RT}} \]
- For the catalyzed reaction (K2):
\[ K2 = A e^{-\frac{EA2}{RT}} = A e^{-\frac{25,000}{RT}} \]

4. **Find the Ratio of Rate Constants**:
- The ratio of the rate constants (K2/K1) is:
\[ \frac{K2}{K1} = \frac{A e^{-\frac{25,000}{RT}}}{A e^{-\frac{100,000}{RT}}} = e^{-\frac{25,000}{RT} + \frac{100,000}{RT}} = e^{\frac{75,000}{RT}} \]

5. **Take the Logarithm**:
- To find \(\log_{10} \left( \frac{K2}{K1} \right)\):
\[ \log_{10} \left( \frac{K2}{K1} \right) = \log_{10} \left( e^{\frac{75,000}{RT}} \right) = \frac{75,000}{RT} \cdot \log_{10}(e) \]
- Since \(\log_{10}(e) \approx 0.4343\), we can write:
\[ \log_{10} \left( \frac{K2}{K1} \right) = \frac{75,000 \cdot 0.4343}{RT} \]

6. **Substitute Values**:
- Given \(R = 8.314 \, \text{J/(mol K)}\) and \(T = 27^{\circ}C = 300 \, K\):
\[ \log_{10} \left( \frac{K2}{K1} \right) = \frac{75,000 \cdot 0.4343}{8.314 \cdot 300} \]

7. **Calculate the Final Value**:
- Calculate the denominator:
\[ RT = 8.314 \cdot 300 = 2494.2 \, \text{J/mol} \]
- Now substitute:
\[ \log_{10} \left( \frac{K2}{K1} \right) = \frac{75,000 \cdot 0.4343}{2494.2} \approx \frac{32,573.5}{2494.2} \approx 13.05 \]

### Final Answer:
\(\log_{10} \left( \frac{K2}{K1} \right) \approx 13.05\)