Conisder the following reaction at `300 K`:
`Ararr B` (uncatalyzed reaction)
`ArarrB` (catalyzed reaction)
The activation energy is lowered by `8.314 kJ mol^(-1)` for the catalyzed reaction. The rate of this reaction is

Options

A. `15` times
B. `38` times
C. `22` times
D. `28` times

Explanation

To solve the problem regarding the rate of the catalyzed reaction compared to the uncatalyzed reaction, we will use the Arrhenius equation and the information provided in the question.

### Step-by-Step Solution:

1. **Understand 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 pre-exponential 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 Activation Energies**:
Let:
- \( E_a \) be the activation energy for the uncatalyzed reaction,
- \( E_{ac} = E_a - 8.314 \, \text{kJ/mol} \) be the activation energy for the catalyzed reaction (since it is lowered by 8.314 kJ/mol).

3. **Write the Rate Constants**:
For the uncatalyzed reaction:
\[ k_1 = A e^{-\frac{E_a}{RT}} \]
For the catalyzed reaction:
\[ k_2 = A e^{-\frac{E_{ac}}{RT}} = A e^{-\frac{E_a - 8.314 \times 10^3}{RT}} \]

4. **Take the Ratio of Rate Constants**:
To find the ratio of the rate constants \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_{ac}}{RT}}}{A e^{-\frac{E_a}{RT}}} = e^{-\frac{E_{ac}}{RT} + \frac{E_a}{RT}} = e^{\frac{8.314 \times 10^3}{RT}} \]

5. **Substitute Values**:
Now substitute \( R = 8.314 \, \text{J/mol·K} \) and \( T = 300 \, \text{K} \):
\[ \frac{k_2}{k_1} = e^{\frac{8.314 \times 10^3}{8.314 \times 300}} = e^{\frac{8.314 \times 10^3}{2494.2}} = e^{3.33} \]

6. **Calculate \( e^{3.33} \)**:
Using a calculator:
\[ e^{3.33} \approx 28.6 \]
Therefore, \( \frac{k_2}{k_1} \approx 28 \).

### Conclusion:
The rate of the catalyzed reaction is approximately 28 times that of the uncatalyzed reaction.