In the case of a first order reaction, the ratio of time required for 99.9% completion to 50% completion is

Options

A. 2
B. 5
C. 10
D. None of these

Answer

To find the ratio of time required for 99.9% completion to that for 50% completion in a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Formula**:
The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
where:
- \( t \) is the time,
- \( k \) is the rate constant,
- \( A \) is the initial concentration,
- \( x \) is the amount of reactant that has reacted.

2. **Calculate Time for 99.9% Completion**:
For 99.9% completion, the amount consumed \( x \) is 99.9% of \( A \). Thus, \( A - x = A - 0.999A = 0.001A \).
Substituting into the formula:
\[ t_{99.9\%} = \frac{2.303}{k} \log \left( \frac{A}{0.001A} \right) = \frac{2.303}{k} \log(1000) \]
Since \( \log(1000) = 3 \):
\[ t_{99.9\%} = \frac{2.303 \times 3}{k} = \frac{6.909}{k} \]

3. **Calculate Time for 50% Completion**:
For 50% completion, the amount consumed \( x \) is 50% of \( A \). Thus, \( A - x = A - 0.5A = 0.5A \).
Substituting into the formula:
\[ t_{50\%} = \frac{2.303}{k} \log \left( \frac{A}{0.5A} \right) = \frac{2.303}{k} \log(2) \]
Since \( \log(2) \approx 0.301 \):
\[ t_{50\%} = \frac{2.303 \times 0.301}{k} \approx \frac{0.693}{k} \]

4. **Find the Ratio of Times**:
Now, we can find the ratio of the times for 99.9% completion to 50% completion:
\[ \text{Ratio} = \frac{t_{99.9\%}}{t_{50\%}} = \frac{\frac{6.909}{k}}{\frac{0.693}{k}} = \frac{6.909}{0.693} \]
Simplifying this gives:
\[ \text{Ratio} \approx 10 \]

### Final Answer:
The ratio of the time required for 99.9% completion to that for 50% completion in a first-order reaction is approximately **10**.