At `227^(@)C` , the presence of catalyst causes the activation energy of a reaction to decrease by `4.606 K cal. `The rate of the reaction will be increased by `: -`

Options

A. 2 times
B. 10 times
C. 100 times
D. 1000 times

Explanation

To solve the problem, we need to determine how the rate of a reaction changes when the activation energy decreases due to the presence of a catalyst. We will use the Arrhenius equation and the relationship between the rate constants and activation energy.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Temperature (T) = 227°C = 500 K (after converting to Kelvin)
- Decrease in activation energy (ΔEa) = 4.606 kcal = 4606 cal (since 1 kcal = 1000 cal)

2. **Use the Arrhenius Equation**:
The relationship between the rate constants (k) and activation energy (Ea) is given by:
\[ \log \frac{k_2}{k_1} = -\frac{\Delta E_a}{2.303RT} \]
where:
- \( R \) = 1.987 cal/(mol·K) (the gas constant)
- \( T \) = 500 K

3. **Substitute the Values**:
Substitute the values into the equation:
\[ \log \frac{k_2}{k_1} = -\frac{4606}{2.303 \times 1.987 \times 500} \]

4. **Calculate the Denominator**:
Calculate the denominator:
\[ 2.303 \times 1.987 \times 500 \approx 2292.5 \]

5. **Calculate the Logarithm**:
Now calculate the logarithm:
\[ \log \frac{k_2}{k_1} = -\frac{4606}{2292.5} \approx -2.009 \]

6. **Convert Logarithm to Ratio**:
To find the ratio \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = 10^{-2.009} \approx 0.0098 \]
However, we need to find \( k_2 \) in terms of \( k_1 \):
\[ k_2 = k_1 \times 10^{2.009} \approx k_1 \times 100 \]

7. **Conclusion**:
Therefore, the rate of the reaction increases by a factor of approximately 100 times due to the catalyst.

### Final Answer:
The rate of the reaction will be increased by **100 times**.