A first order reaction takes `69.3` minutes for `50%` completion. How much time will be needed for `80%` completion?

Explanation

To solve the problem, we need to determine the time required for an 80% completion of a first-order reaction, given that the time for 50% completion (half-life, \( t_{1/2} \)) is 69.3 minutes.

### Step-by-Step Solution:

1. **Identify the half-life of the reaction**:
The half-life (\( t_{1/2} \)) for the reaction is given as 69.3 minutes.

2. **Calculate the rate constant \( k \)**:
For a first-order reaction, the relationship between half-life and the rate constant is given by:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the value of \( t_{1/2} \):
\[ k = \frac{0.693}{69.3} \approx 0.01 \, \text{min}^{-1} \]

3. **Use the first-order kinetics equation**:
The first-order kinetics equation is:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
where:
- \( A \) is the initial concentration,
- \( x \) is the amount reacted.

For 80% completion, \( A - x = 20\% \) of \( A \). Therefore, if we take \( A = 100 \) (for simplicity), then \( A - x = 20 \).

4. **Substituting values into the equation**:
Now substituting \( A = 100 \), \( A - x = 20 \), and \( k \):
\[ t = \frac{2.303}{0.01} \log \left( \frac{100}{20} \right) \]

5. **Calculate the logarithm**:
\[ \log \left( \frac{100}{20} \right) = \log(5) \approx 0.699 \]

6. **Calculate the time \( t \)**:
Now substituting the value of the logarithm:
\[ t = \frac{2.303}{0.01} \times 0.699 \approx 230.3 \, \text{minutes} \]

7. **Final calculation**:
\[ t \approx 230.3 \, \text{minutes} \]

### Conclusion:
The time needed for 80% completion of the reaction is approximately **230.3 minutes**.