Decomposition of a gas is of first order. It takes 80 minutes for `80%` of the gas to be decomposed when its initial concentration is `8xx10^(-2)` mole/litre. Calculate the specific reaction rate.

Explanation

To solve the problem, we need to calculate the specific reaction rate (rate constant, \( k \)) for the first-order decomposition of a gas. We know the following:

- Initial concentration, \( [A]_0 = 8 \times 10^{-2} \) mol/L
- Time taken for 80% decomposition, \( t = 80 \) minutes
- Percentage decomposed = 80%, thus the concentration remaining, \( [A]_t = 20\% \) of \( [A]_0 \)

### Step-by-Step Solution:

1. **Calculate the concentration remaining after 80% decomposition**:
\[ [A]_t = [A]_0 \times (1 - 0.80) = 8 \times 10^{-2} \times 0.20 = 8 \times 10^{-2} \times \frac{2}{10} = 1.6 \times 10^{-2} \text{ mol/L} \]

2. **Use the first-order rate equation**:
The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A]_0}{[A]_t} \right) = k \cdot t \]

3. **Substituting the known values into the equation**:
\[ \ln \left( \frac{8 \times 10^{-2}}{1.6 \times 10^{-2}} \right) = k \cdot 80 \]

4. **Calculate the ratio**:
\[ \frac{8 \times 10^{-2}}{1.6 \times 10^{-2}} = 5 \]

5. **Calculate the natural logarithm**:
\[ \ln(5) \approx 1.6094 \]

6. **Substituting back into the equation**:
\[ 1.6094 = k \cdot 80 \]

7. **Solve for \( k \)**:
\[ k = \frac{1.6094}{80} \approx 0.0201 \text{ min}^{-1} \]

### Final Answer:
The specific reaction rate \( k \) is approximately \( 0.0201 \text{ min}^{-1} \).