For a first order reaction, the ratio of time for the completion of `99.9%` and half of the reaction is

Options

A. `8`
B. `10`
C. `9`
D. `12`

Explanation

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

### Step 1: Understand the first-order reaction kinetics
For a first-order reaction, the time taken to reach a certain percentage of completion can be calculated using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{A_0}{A - x} \right) \]
where:
- \( T \) is the time,
- \( k \) is the rate constant,
- \( A_0 \) is the initial concentration,
- \( A \) is the concentration at time \( T \),
- \( x \) is the amount reacted.

### Step 2: Calculate \( T_{1/2} \) (time for half the reaction)
For half the reaction, \( x = \frac{A_0}{2} \):
\[ T_{1/2} = \frac{2.303}{k} \log \left( \frac{A_0}{A_0 - \frac{A_0}{2}} \right) \]
This simplifies to:
\[ T_{1/2} = \frac{2.303}{k} \log \left( \frac{A_0}{\frac{A_0}{2}} \right) = \frac{2.303}{k} \log(2) \]

### Step 3: Calculate \( T_{99.9\%} \) (time for 99.9% completion)
For 99.9% completion, \( x = 0.999 A_0 \):
\[ T_{99.9\%} = \frac{2.303}{k} \log \left( \frac{A_0}{A_0 - 0.999 A_0} \right) \]
This simplifies to:
\[ T_{99.9\%} = \frac{2.303}{k} \log \left( \frac{A_0}{0.001 A_0} \right) = \frac{2.303}{k} \log(1000) = \frac{2.303}{k} \cdot 3 \]

### Step 4: Form the ratio \( \frac{T_{99.9\%}}{T_{1/2}} \)
Now we can find the ratio:
\[ \frac{T_{99.9\%}}{T_{1/2}} = \frac{\frac{2.303}{k} \cdot 3}{\frac{2.303}{k} \log(2)} \]
The \( \frac{2.303}{k} \) cancels out:
\[ \frac{T_{99.9\%}}{T_{1/2}} = \frac{3}{\log(2)} \]

### Step 5: Calculate the numerical value
Using the value \( \log(2) \approx 0.301 \):
\[ \frac{T_{99.9\%}}{T_{1/2}} \approx \frac{3}{0.301} \approx 9.965 \]

### Conclusion
Thus, the ratio of the time for the completion of 99.9% of the reaction to the time for the completion of half of the reaction is approximately **10**.

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