Calculate the `log_(10)` of the ratio of the catalysed and uncatalysed rate constant at `25^(@)`C. If at this temperature the energy of activation of a catalysed reaction is 162KJ and for the uncatalysed reaction the value is 350 KJ
(Assume frequency factor is same for both reaction)

Options

A. 28.94
B. 32.94
C. 36.94
D. 40.94

Explanation

To solve the problem, we need to calculate the logarithm (base 10) of the ratio of the rate constants for a catalyzed reaction and an uncatalyzed reaction at 25°C. We will use the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea) and temperature (T).

### Step-by-Step Solution:

1. **Write the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the frequency factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

2. **Define the Rate Constants**:
Let:
- \( k \) be the rate constant for the catalyzed reaction (with catalyst),
- \( k' \) be the rate constant for the uncatalyzed reaction (without catalyst).

For the catalyzed reaction:
\[ k = A e^{-\frac{E_{a, \text{catalyzed}}}{RT}} = A e^{-\frac{162 \times 10^3}{8.314 \times 298}} \]
For the uncatalyzed reaction:
\[ k' = A e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}} = A e^{-\frac{350 \times 10^3}{8.314 \times 298}} \]

3. **Calculate the Ratio of Rate Constants**:
The ratio of the rate constants is:
\[ \frac{k}{k'} = \frac{A e^{-\frac{E_{a, \text{catalyzed}}}{RT}}}{A e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}}} \]
The frequency factor \( A \) cancels out:
\[ \frac{k}{k'} = e^{-\frac{E_{a, \text{catalyzed}} - E_{a, \text{uncatalyzed}}}{RT}} \]

4. **Substitute Values**:
Substituting the values of activation energies:
\[ E_{a, \text{uncatalyzed}} = 350 \, \text{kJ} = 350 \times 10^3 \, \text{J} \]
\[ E_{a, \text{catalyzed}} = 162 \, \text{kJ} = 162 \times 10^3 \, \text{J} \]
Thus,
\[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = (350 - 162) \times 10^3 = 188 \times 10^3 \, \text{J} \]

5. **Calculate the Exponent**:
Now we can calculate:
\[ \frac{k}{k'} = e^{-\frac{188 \times 10^3}{8.314 \times 298}} \]

6. **Calculate the Exponent Value**:
First, calculate the denominator:
\[ 8.314 \times 298 \approx 2477.572 \]
Now calculate:
\[ -\frac{188000}{2477.572} \approx -75.8 \]
Therefore:
\[ \frac{k}{k'} = e^{-75.8} \]

7. **Calculate the Logarithm**:
Now we need to find:
\[ \log_{10}\left(\frac{k}{k'}\right) = \log_{10}(e^{-75.8}) = -75.8 \times \log_{10}(e) \]
Using \(\log_{10}(e) \approx 0.4343\):
\[ \log_{10}\left(\frac{k}{k'}\right) \approx -75.8 \times 0.4343 \approx -32.9 \]

### Final Answer:
\[ \log_{10}\left(\frac{k}{k'}\right) \approx -32.9 \]