In a reaction carried out at `500 K, 0.001%` of the total number of colliisons are effective. The energy of activation of the reaction is approximately

Options

A. `15.8 kcal mol^(-1)`
B. `11.5 kcal mol^(-1)`
C. `12.8 kcal mol^(-1)`
D. zero

Explanation

To solve the problem, we need to determine the activation energy (Ea) of a reaction given that at 500 K, 0.001% of the total number of collisions are effective. We will use the Arrhenius equation which relates the fraction of effective collisions to the activation energy.

### Step-by-Step Solution:

1. **Convert the percentage of effective collisions to a fraction**:
\[ x = \frac{0.001}{100} = 0.00001 = 10^{-5} \]

2. **Use the Arrhenius equation**:
The fraction of molecules that have enough energy to react is given by:
\[ x = e^{-\frac{E_a}{RT}} \]
where:
- \( x \) is the fraction of effective collisions,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

3. **Substitute known values**:
We need to express \( R \) in calories. The value of \( R \) is approximately \( 2 \, \text{cal/mol·K} \).
\[ T = 500 \, \text{K} \]

4. **Take the natural logarithm of both sides**:
\[ \ln(x) = -\frac{E_a}{RT} \]
Substituting \( x = 10^{-5} \):
\[ \ln(10^{-5}) = -\frac{E_a}{(2)(500)} \]

5. **Calculate \( \ln(10^{-5}) \)**:
\[ \ln(10^{-5}) = -5 \ln(10) \quad \text{(since } \ln(a^b) = b \ln(a)\text{)} \]
Using \( \ln(10) \approx 2.303 \):
\[ \ln(10^{-5}) = -5 \times 2.303 = -11.515 \]

6. **Substitute back into the equation**:
\[ -11.515 = -\frac{E_a}{(2)(500)} \]

7. **Solve for \( E_a \)**:
Rearranging gives:
\[ E_a = 11.515 \times (2)(500) \]
\[ E_a = 11.515 \times 1000 = 11515 \, \text{calories} \]

8. **Convert to kilocalories**:
\[ E_a = \frac{11515}{1000} = 11.515 \, \text{kcal/mol} \]

### Final Answer:
The activation energy \( E_a \) is approximately **11.5 kcal/mol**.