For a first order reaction, if the time taken for completion of `50%` of the reaction is `t` second, the time required for completion of `99.9%` of the reaction is `nt`. Find the value of n?

Answer

To solve the problem, we will use the integrated rate law for first-order reactions and the concept of half-life. Let's break down the solution step by step.

### Step 1: Understand the first-order reaction
For a first-order reaction, the integrated rate equation is given by:
\[ \ln \left( \frac{A_0}{A} \right) = k t \]
where:
- \( A_0 \) = initial concentration
- \( A \) = concentration at time \( t \)
- \( k \) = rate constant
- \( t \) = time

### Step 2: Determine the time for 50% completion
For 50% completion, we have:
- \( A_0 = 100 \) (initial concentration)
- \( A = 50 \) (concentration after 50% completion)

Substituting these values into the integrated rate equation:
\[ \ln \left( \frac{100}{50} \right) = k t \]
This simplifies to:
\[ \ln(2) = k t \]
Thus, we can express \( t \) as:
\[ t = \frac{\ln(2)}{k} \]

### Step 3: Determine the time for 99.9% completion
For 99.9% completion, we have:
- \( A_0 = 100 \)
- \( A = 0.1 \) (concentration after 99.9% completion)

Substituting these values into the integrated rate equation:
\[ \ln \left( \frac{100}{0.1} \right) = k (nt) \]
This simplifies to:
\[ \ln(1000) = k (nt) \]
Thus, we can express \( nt \) as:
\[ nt = \frac{\ln(1000)}{k} \]

### Step 4: Relate the two expressions for time
We have two expressions:
1. \( t = \frac{\ln(2)}{k} \)
2. \( nt = \frac{\ln(1000)}{k} \)

We can substitute the expression for \( t \) from the first equation into the second:
\[ n \left( \frac{\ln(2)}{k} \right) = \frac{\ln(1000)}{k} \]

### Step 5: Cancel \( k \) and solve for \( n \)
Cancelling \( k \) from both sides gives:
\[ n \ln(2) = \ln(1000) \]
Now, we know that:
\[ \ln(1000) = \ln(10^3) = 3 \ln(10) \]
Thus, we can write:
\[ n \ln(2) = 3 \ln(10) \]
Now, solving for \( n \):
\[ n = \frac{3 \ln(10)}{\ln(2)} \]

### Step 6: Calculate the value of \( n \)
Using the approximate values:
- \( \ln(10) \approx 2.303 \)
- \( \ln(2) \approx 0.693 \)

We can substitute these values:
\[ n \approx \frac{3 \times 2.303}{0.693} \approx \frac{6.909}{0.693} \approx 9.95 \approx 10 \]

### Final Answer
Thus, the value of \( n \) is approximately:
\[ \boxed{10} \]