The time required for the decompoistion of `99.9%` fraction of a first order reaction is………….to that of its half-life time.

Answer

To solve the problem, we need to determine the relationship between the time required for the decomposition of 99.9% of a substance in a first-order reaction and its half-life time.

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Kinetics**:
The time required for a first-order reaction can be calculated using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
where:
- \( T \) is the time taken for the reaction.
- \( k \) is the rate constant.
- \( [A_0] \) is the initial concentration.
- \( [A] \) is the concentration at time \( T \).

2. **Calculate Half-Life**:
For the half-life (\( T_{1/2} \)), the remaining concentration after half-life is half of the initial concentration:
\[ [A] = \frac{[A_0]}{2} \]
Substituting this into the formula gives:
\[ T_{1/2} = \frac{2.303}{k} \log \left( \frac{[A_0]}{\frac{[A_0]}{2}} \right) = \frac{2.303}{k} \log(2) \]
Since \( \log(2) \approx 0.3010 \):
\[ T_{1/2} = \frac{2.303 \times 0.3010}{k} \]

3. **Calculate Time for 99.9% Decomposition**:
For 99.9% decomposition, the remaining concentration will be 0.1% of the initial concentration:
\[ [A] = 0.1\% \text{ of } [A_0] = 0.001 [A_0] \]
Substituting this into the formula gives:
\[ T_{99.9\%} = \frac{2.303}{k} \log \left( \frac{[A_0]}{0.001 [A_0]} \right) = \frac{2.303}{k} \log(1000) \]
Since \( \log(1000) = 3 \):
\[ T_{99.9\%} = \frac{2.303 \times 3}{k} \]

4. **Relate \( T_{99.9\%} \) to \( T_{1/2} \)**:
Now we can find the ratio of \( T_{99.9\%} \) to \( T_{1/2} \):
\[ \frac{T_{99.9\%}}{T_{1/2}} = \frac{\frac{2.303 \times 3}{k}}{\frac{2.303 \times 0.3010}{k}} = \frac{3}{0.3010} \]
Calculating this gives:
\[ \frac{T_{99.9\%}}{T_{1/2}} \approx 9.965 \approx 10 \]

5. **Conclusion**:
Therefore, the time required for the decomposition of 99.9% of a substance in a first-order reaction is approximately **10 times** that of its half-life.