For the first order reaction `ArarrB+C` , carried out at `27^(@)C` if `3.8xx10^(-16)%` of the reactant molecules exists in the activated state, the `E_(a)` (activation energy) of the reaction is :

Options

A. `12kJ//mol`
B. `831.4kJ//mol`
C. `100kJ//mol`
D. `88.57kJ//mol`

Answer

To solve the problem of finding the activation energy (E_a) for the first-order reaction \( A \rightarrow B + C \) at \( 27^\circ C \), we will use the Arrhenius equation and the given percentage of molecules in the activated state.

### 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. **Express the Fraction of Activated Molecules**:
We know that the fraction of molecules in the activated state can be expressed as:
\[ \frac{k}{A} = e^{-\frac{E_a}{RT}} \]
Given that \( 3.8 \times 10^{-16} \% \) of the reactant molecules exist in the activated state, we convert this percentage to a fraction:
\[ \frac{k}{A} = \frac{3.8 \times 10^{-16}}{100} = 3.8 \times 10^{-18} \]

3. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{k}{A}\right) = -\frac{E_a}{RT} \]
Substituting the value we found:
\[ \ln(3.8 \times 10^{-18}) = -\frac{E_a}{RT} \]

4. **Calculate the Temperature in Kelvin**:
The temperature given is \( 27^\circ C \). To convert this to Kelvin:
\[ T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \]

5. **Substitute Values into the Equation**:
We know \( R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \). Now we can substitute \( R \) and \( T \) into the equation:
\[ \ln(3.8 \times 10^{-18}) = -\frac{E_a}{(0.008314)(300)} \]

6. **Calculate the Natural Logarithm**:
Calculate \( \ln(3.8 \times 10^{-18}) \):
\[ \ln(3.8 \times 10^{-18}) \approx -39.4 \]

7. **Rearranging the Equation**:
Rearranging the equation to solve for \( E_a \):
\[ -39.4 = -\frac{E_a}{(0.008314)(300)} \]
\[ E_a = 39.4 \times (0.008314)(300) \]

8. **Calculate \( E_a \)**:
Now calculate \( E_a \):
\[ E_a = 39.4 \times 2.4942 \approx 98.5 \text{ kJ/mol} \]

9. **Final Value**:
Rounding this value gives us approximately \( 100 \text{ kJ/mol} \).

### Conclusion:
The activation energy \( E_a \) for the reaction is approximately \( 100 \text{ kJ/mol} \).