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 solve the problem of finding 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 1: Understand the formula for time in a first-order reaction
For a first-order reaction, the time required for a certain percentage completion can be calculated using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{A}{A - X} \right) \]
where:
- \( T \) is the time,
- \( k \) is the rate constant,
- \( A \) is the initial concentration,
- \( X \) is the amount reacted.

### Step 2: Calculate time for 99.9% completion
For 99.9% completion, we have:
- Initial concentration \( A = 100 \) (assuming 100%),
- Amount reacted \( X = 99.9 \).

Thus, the remaining concentration after 99.9% completion is:
\[ A - X = 100 - 99.9 = 0.1 \]

Now, substituting into the formula:
\[ T_{99.9\%} = \frac{2.303}{k} \log \left( \frac{100}{0.1} \right) \]
\[ = \frac{2.303}{k} \log (1000) \]
\[ = \frac{2.303}{k} \cdot 3 \]
\[ = \frac{6.909}{k} \]

### Step 3: Calculate time for 50% completion
For 50% completion, we have:
- Amount reacted \( X = 50 \).

Thus, the remaining concentration after 50% completion is:
\[ A - X = 100 - 50 = 50 \]

Now, substituting into the formula:
\[ T_{50\%} = \frac{2.303}{k} \log \left( \frac{100}{50} \right) \]
\[ = \frac{2.303}{k} \log (2) \]
\[ = \frac{2.303}{k} \cdot 0.301 \]
\[ = \frac{0.69303}{k} \]

### Step 4: Find the ratio of times
Now, we need to find the ratio of \( T_{99.9\%} \) to \( T_{50\%} \):
\[ \text{Ratio} = \frac{T_{99.9\%}}{T_{50\%}} = \frac{\frac{6.909}{k}}{\frac{0.69303}{k}} \]

The \( k \) cancels out:
\[ = \frac{6.909}{0.69303} \approx 9.96 \]

### Step 5: Conclusion
The ratio of time required for 99.9% completion to that for 50% completion is approximately equal to 10.

Thus, the answer is:
**10**

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