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 the reaction
B. 100 times that required for two-third 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 solve the problem of determining the time required for 99.9% of a first-order reaction to take place, we can follow these steps:

### Step 1: Understand the Reaction Order
For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]

where:
- \( t \) is the time taken for the reaction,
- \( k \) is the rate constant,
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \).

### Step 2: Define the Concentrations for 99.9% Completion
If 99.9% of the reaction has taken place, then 0.1% of the reactant remains. This means:

- \( [A_0] = 1 \) (initial concentration)
- \( [A] = 0.001 \) (final concentration after 99.9% reaction)

### Step 3: Substitute Values into the Integrated Rate Law
Now, substituting these values into the integrated rate law:

\[ t_{99.9\%} = \frac{2.303}{k} \log \left( \frac{1}{0.001} \right) \]

### Step 4: Calculate the Logarithm
Calculating the logarithm:

\[ \log \left( \frac{1}{0.001} \right) = \log(1000) = 3 \]

### Step 5: Substitute Back into the Equation
Now, substitute this back into the equation for \( t_{99.9\%} \):

\[ t_{99.9\%} = \frac{2.303}{k} \times 3 \]

This simplifies to:

\[ t_{99.9\%} = \frac{6.909}{k} \]

### Step 6: Relate to Half-Life
The half-life (\( t_{1/2} \)) for a first-order reaction is given by:

\[ t_{1/2} = \frac{0.693}{k} \]

### Step 7: Find the Ratio of \( t_{99.9\%} \) to \( t_{1/2} \)
Now, we can find the ratio of \( t_{99.9\%} \) to \( t_{1/2} \):

\[ \frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909/k}{0.693/k} = \frac{6.909}{0.693} \]

### Step 8: Calculate the Final Ratio
Calculating this gives:

\[ \frac{t_{99.9\%}}{t_{1/2}} \approx 10 \]

### Conclusion
Thus, the time required for 99.9% of the reaction to take place is nearly 10 times that required for the half of the reaction.

### Final Answer
**The correct option is:** 10 times that required for the half of the reaction.

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