Show that in case of a first order reaction, the time required for 99.9% of the reaction to take place is about ten times than that required for half the reaction.

Explanation

To show that in the case of a first-order reaction, the time required for 99.9% of the reaction to take place is about ten times that required for half the reaction, we can follow these steps:

### Step 1: Understand the First-Order Reaction
A first-order reaction can be represented as:
\[ A \rightarrow P \]
The rate of the reaction is given by:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of reactant \( A \).

### Step 2: Write the Integrated Rate Law for First-Order Reactions
The integrated rate law for a first-order reaction is:
\[ \ln \left( \frac{[A_0]}{[A_t]} \right) = kt \]
where:
- \( [A_0] \) is the initial concentration,
- \( [A_t] \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

### Step 3: Calculate the Half-Life (\( t_{1/2} \))
The half-life for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]

### Step 4: Calculate the Time for 99.9% Completion
For 99.9% of the reaction to take place, 0.1% of the reactant remains. If we assume the initial concentration \( [A_0] = 100 \), then:
\[ [A_t] = 0.1 \]
Now, substituting into the integrated rate law:
\[ \ln \left( \frac{100}{0.1} \right) = kt_{99.9\%} \]
This simplifies to:
\[ \ln(1000) = kt_{99.9\%} \]

### Step 5: Simplify \( \ln(1000) \)
We can express \( \ln(1000) \) as:
\[ \ln(1000) = \ln(10^3) = 3 \ln(10) \]
Using the approximation \( \ln(10) \approx 2.303 \):
\[ \ln(1000) \approx 3 \times 2.303 = 6.909 \]

### Step 6: Substitute Back to Find \( t_{99.9\%} \)
Now, we can express \( t_{99.9\%} \):
\[ t_{99.9\%} = \frac{6.909}{k} \]

### Step 7: Relate \( t_{99.9\%} \) to \( t_{1/2} \)
We know that:
\[ t_{1/2} = \frac{0.693}{k} \]
Now, we can express \( t_{99.9\%} \) in terms of \( t_{1/2} \):
\[ t_{99.9\%} = \frac{6.909}{k} = 10 \times \frac{0.693}{k} = 10 \times t_{1/2} \]

### Conclusion
Thus, we have shown that the time required for 99.9% of the reaction to take place is approximately ten times that required for half the reaction:
\[ t_{99.9\%} \approx 10 \times t_{1/2} \]