For a first order reaction, the time required for `99.9%` of the reaction to take place is nearly

Options

A. 10 times that required for half of the reaction
B. 100 times that required for two-thirds of the reaction
C. 10 times that required for one-fourth of the reaction
D. 20 times that required for half of the reaction

Answer

To determine the time required for 99.9% of a first-order reaction to take place, we can follow these steps:

### Step 1: Understand the concept of first-order reactions
In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction can be expressed as:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
where:
- \([A]_0\) is the initial concentration,
- \([A]\) is the concentration at time \(t\),
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 2: Calculate the fraction of the reaction that has occurred
If 99.9% of the reaction has taken place, then the remaining concentration \([A]\) is:
\[ [A] = [A]_0 \times (1 - 0.999) = [A]_0 \times 0.001 \]

### Step 3: Substitute into the integrated rate law
Substituting \([A]\) into the integrated rate law gives:
\[ \ln \left( \frac{[A]_0}{[A]_0 \times 0.001} \right) = kt \]
This simplifies to:
\[ \ln(1000) = kt \]

### Step 4: Calculate \(\ln(1000)\)
Using the properties of logarithms:
\[ \ln(1000) = \ln(10^3) = 3 \ln(10) \]
Given that \(\ln(10) \approx 2.303\):
\[ \ln(1000) \approx 3 \times 2.303 \approx 6.909 \]

### Step 5: Relate time to half-life
For a first-order reaction, the half-life (\(t_{1/2}\)) is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
The time required for 99.9% completion is approximately 10 times the half-life:
\[ t \approx 10 \times t_{1/2} = 10 \times \frac{0.693}{k} \]

### Step 6: Final expression for time
Thus, the time required for 99.9% of the reaction to occur is:
\[ t \approx 10 \times \frac{0.693}{k} \approx \frac{6.93}{k} \]

### Conclusion
The time required for 99.9% of the reaction to take place is nearly 10 times the half-life of the reaction.

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