The rate constant for the decomposition of a certain substance is `2.80 xx 10^(-1)M^(-1)s^(-1)` at `30^(@)C` and `1.38 xx 10^(-2)M^(-1) s^(-1)` at `50^(@)C`. The Arrhenius parameters (A) of the reaction is:
(`R=8.314 xx 10^(-3) kJ mol^(-1) K^(-1)`).

Options

A. `8.68 xx 10^(8) M^(-1) s^(-1)`
B. `2.16 xx 10^(7) M^(-1) s^(-1)`
C. `4.34 xx 10^(8) M^(-1) s^(-1)`
D. `3.34 xx 10^(8) M^(-1) s^(-1)`

Explanation

To solve for the Arrhenius parameter \( A \) given the rate constants at two different temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Rate constant at \( T_1 = 30^\circ C = 303 \, K \): \( K_1 = 2.80 \times 10^{-1} \, M^{-1}s^{-1} \)
- Rate constant at \( T_2 = 50^\circ C = 323 \, K \): \( K_2 = 1.38 \times 10^{-2} \, M^{-1}s^{-1} \)
- Universal gas constant \( R = 8.314 \times 10^{-3} \, kJ \, mol^{-1} \, K^{-1} \)

2. **Use the Arrhenius Equation:**
The Arrhenius equation can be expressed in logarithmic form for two temperatures:
\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Rearranging gives us:
\[ E_a = -R \cdot \frac{\ln \left( \frac{K_2}{K_1} \right)}{\left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \]

3. **Calculate \( \ln \left( \frac{K_2}{K_1} \right) \):**
\[ \frac{K_2}{K_1} = \frac{1.38 \times 10^{-2}}{2.80 \times 10^{-1}} = 0.0492857 \]
\[ \ln \left( 0.0492857 \right) \approx -3.006 \]

4. **Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):**
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{323} - \frac{1}{303} = \frac{303 - 323}{323 \times 303} = \frac{-20}{97959} \approx -0.000204 \]

5. **Substitute Values into the Equation for \( E_a \):**
\[ E_a = - (8.314 \times 10^{-3}) \cdot \frac{-3.006}{-0.000204} \]
\[ E_a \approx 64.91 \, kJ/mol \]

6. **Use the Arrhenius Equation to Find \( A \):**
Rearranging the Arrhenius equation gives:
\[ A = \frac{K_1}{e^{-\frac{E_a}{RT_1}}} \]
Substitute \( K_1 \), \( E_a \), \( R \), and \( T_1 \):
\[ A = \frac{2.80 \times 10^{-1}}{e^{-\frac{64910}{8.314 \times 303}}} \]
Calculate \( e^{-\frac{64910}{8.314 \times 303}} \):
\[ \frac{64910}{8.314 \times 303} \approx 25.68 \implies e^{-25.68} \approx 0 \]
Thus, \( A \) can be approximated as:
\[ A \approx 2.80 \times 10^{-1} \cdot e^{25.68} \]
Calculate \( e^{25.68} \approx 1.59 \times 10^{11} \):
\[ A \approx 2.80 \times 10^{-1} \cdot 1.59 \times 10^{11} \approx 4.45 \times 10^{10} \, M^{-1}s^{-1} \]

### Final Answer:
The Arrhenius parameter \( A \) is approximately \( 4.45 \times 10^{10} \, M^{-1}s^{-1} \).