The ratio of the times of 99.9 % of the reaction to complete and half of the reaction to complete is

Options

A. `2`
B. `4`
C. `8`
D. `10`

Explanation

To solve the problem of finding the ratio of the times of 99.9% completion of a first-order reaction to the time of 50% completion, we can follow these steps:

### Step-by-Step Solution:

1. **Assume the Reaction Order**:
We assume that the reaction is a first-order reaction.

2. **Use the First-Order Reaction Equation**:
The time required for a first-order reaction to reach a certain concentration can be described by the equation:
\[ T = \frac{2.303}{k} \log \left( \frac{C_0}{C_t} \right) \]
where:
- \( T \) is the time,
- \( k \) is the rate constant,
- \( C_0 \) is the initial concentration,
- \( C_t \) is the concentration at time \( T \).

3. **Define the Concentrations for 99.9% and 50% Completion**:
- For 99.9% completion, \( C_t \) will be 0.1% of \( C_0 \). If we assume \( C_0 = 100 \, \text{M} \), then:
\[ C_1 = 0.1 \, \text{M} \]
- For 50% completion, \( C_t \) will be 50% of \( C_0 \):
\[ C_2 = 50 \, \text{M} \]

4. **Calculate the Times**:
- For 99.9% completion:
\[ T_{99.9} = \frac{2.303}{k} \log \left( \frac{100}{0.1} \right) = \frac{2.303}{k} \log(1000) = \frac{2.303}{k} \cdot 3 \]
- For 50% completion:
\[ T_{50} = \frac{2.303}{k} \log \left( \frac{100}{50} \right) = \frac{2.303}{k} \log(2) \]

5. **Set Up the Ratio**:
\[ \frac{T_{99.9}}{T_{50}} = \frac{\frac{2.303 \cdot 3}{k}}{\frac{2.303 \cdot \log(2)}{k}} = \frac{3}{\log(2)} \]

6. **Calculate the Value of \(\log(2)\)**:
The approximate value of \(\log(2)\) is 0.301. Thus:
\[ \frac{T_{99.9}}{T_{50}} \approx \frac{3}{0.301} \approx 9.965 \]

7. **Final Result**:
The ratio \( \frac{T_{99.9}}{T_{50}} \) is approximately equal to 10.

### Conclusion:
The ratio of the times of 99.9% of the reaction to complete and half of the reaction to complete is approximately **10**.

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